NCERT Solutions for Class 12 Maths Chapter 4 Determinants
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Page 1
NCERT
SOLUTIONS
CLASS - 12th
aglase .co
SOLUTIONS
CLASS - 12th
aglase .co
Page 2
Class : 12th
Subject : Maths
Chapter : 4
Chapter Name : Determinants
Q1 Evaluate the determinants In Exercises 1 and 2
| 2
−5
4
−1 |
Answer.
| 2
−5
4
−1 |
= 2(-1) -4(-5) = -2 + 20 = 18
Page : 108 , Block Name : Exercise 4.1
Q2
(i) | cosθ
sinθ
− sinθ
cosθ |
(ii)
x+1|
x2 − x + 1 x−1
x+1 |
Answer.
| cosθ
sinθ
− sinθ
cosθ |
(cosθ)(cosθ) − ( − sinθ)(sinθ) = cos 2θ + sin 2θ = 1
(ii)
| x2 − x + 1
x+1
x−1
x+1 |
( )
= x 2 − x + 1 (x + 1) − (x − 1)(x + 1)
= x3 − x2 + x + x2 − x + 1 − x2 − 1 ( )
= x3 + 1 − x2 + 1
= x3 − x2 + 2
Subject : Maths
Chapter : 4
Chapter Name : Determinants
Q1 Evaluate the determinants In Exercises 1 and 2
| 2
−5
4
−1 |
Answer.
| 2
−5
4
−1 |
= 2(-1) -4(-5) = -2 + 20 = 18
Page : 108 , Block Name : Exercise 4.1
Q2
(i) | cosθ
sinθ
− sinθ
cosθ |
(ii)
x+1|
x2 − x + 1 x−1
x+1 |
Answer.
| cosθ
sinθ
− sinθ
cosθ |
(cosθ)(cosθ) − ( − sinθ)(sinθ) = cos 2θ + sin 2θ = 1
(ii)
| x2 − x + 1
x+1
x−1
x+1 |
( )
= x 2 − x + 1 (x + 1) − (x − 1)(x + 1)
= x3 − x2 + x + x2 − x + 1 − x2 − 1 ( )
= x3 + 1 − x2 + 1
= x3 − x2 + 2
Page 3
Page : 108 , Block Name : Exercise 4.1
Q3 If A= [ ] 1
4
2
2
, then show that | 2A | = 4 | A |
Answer. The given matrix is
A=
[ ]1
4
2
2
[ ] [ ]
∴ 2A = 2
1
4
2
2
=
2
8
4
4
| |
∴ L. L. H. S. = | 2A | =
2
8
4
4
= 2 × 4 − 4 × 8 = 8 − 32 = − 24
| |
Now | A | =
1
4
2
2
=1×2−2×4=2−8= −6
∴ R. H. S. = 4 | A | = 4 × ( − 6) = − 24
∴ L. H. S. = R. H. S.
Page : 108 , Block Name : Exercise 4.1
[ ]
1 0 1
Q4 If A= 0 1 2 , then show that | 3A | = 27 | A |
0 0 4
[ ]
1 0 1
Answer. The given matrix is A = 0 1 2
0 0 4
It can be observed that in the rst column, two entries are zero.
Thus,we expand along the rst column (C1) for easier calculation.
|A| = 1 | | | | | |
1 2
0 4
−0
0 1
0 4
+0
0
1
1
2
= 1(4 − 0) − 0 + 0 = 4
∴ 27 | A | = 27(4) = 108 …(i)
[ ][ ]
1 0 1 3 0 3
Now 3A = 3 0 1 2 = 0 3 6
0 0 4 0 0 12
Q3 If A= [ ] 1
4
2
2
, then show that | 2A | = 4 | A |
Answer. The given matrix is
A=
[ ]1
4
2
2
[ ] [ ]
∴ 2A = 2
1
4
2
2
=
2
8
4
4
| |
∴ L. L. H. S. = | 2A | =
2
8
4
4
= 2 × 4 − 4 × 8 = 8 − 32 = − 24
| |
Now | A | =
1
4
2
2
=1×2−2×4=2−8= −6
∴ R. H. S. = 4 | A | = 4 × ( − 6) = − 24
∴ L. H. S. = R. H. S.
Page : 108 , Block Name : Exercise 4.1
[ ]
1 0 1
Q4 If A= 0 1 2 , then show that | 3A | = 27 | A |
0 0 4
[ ]
1 0 1
Answer. The given matrix is A = 0 1 2
0 0 4
It can be observed that in the rst column, two entries are zero.
Thus,we expand along the rst column (C1) for easier calculation.
|A| = 1 | | | | | |
1 2
0 4
−0
0 1
0 4
+0
0
1
1
2
= 1(4 − 0) − 0 + 0 = 4
∴ 27 | A | = 27(4) = 108 …(i)
[ ][ ]
1 0 1 3 0 3
Now 3A = 3 0 1 2 = 0 3 6
0 0 4 0 0 12
Page 4
∴ | 3A | = 3 | | | | | |
3
0
6
12
−0
0
0
3
12
+0
0
3
3
6
= 3(36 − 0) = 3(36) = 108…(ii)
From equation (i) and (ii), we have :
|3 A|=27|A|
Hence, the given result is proved.
Page : 108 , Block Name : Exercise 4.1
Q5 Evaluate the determinants
| |
3 −1 −2
(i) A = 0 0 −1
3 −5 0
| |
3 −4 5
(ii) 1 1 −2
2 3 1
| |
0 1 2
(iii) − 1 0 −3
−2 3 0
| |
2 −1 −2
(iv) 0 2 −1
3 −5 0
Answer.
| |
3 −1 −2
(i) Let A = 0 0 −1
3 −5 0
It can be observed that in the second row,two entries are zero.
Thus, we expand along the second row for easier calculation.
|A| = − 0
| −1
−5
−2
0 | | |
+0
3
3
−2
0
− ( − 1)
| |
3
3
−1
−5
= ( − 15 + 3) = − 12
| |
3 −4 5
(ii) Let 1 1 −2
2 3 1
By expanding along the rst row, we have :
3
0
6
12
−0
0
0
3
12
+0
0
3
3
6
= 3(36 − 0) = 3(36) = 108…(ii)
From equation (i) and (ii), we have :
|3 A|=27|A|
Hence, the given result is proved.
Page : 108 , Block Name : Exercise 4.1
Q5 Evaluate the determinants
| |
3 −1 −2
(i) A = 0 0 −1
3 −5 0
| |
3 −4 5
(ii) 1 1 −2
2 3 1
| |
0 1 2
(iii) − 1 0 −3
−2 3 0
| |
2 −1 −2
(iv) 0 2 −1
3 −5 0
Answer.
| |
3 −1 −2
(i) Let A = 0 0 −1
3 −5 0
It can be observed that in the second row,two entries are zero.
Thus, we expand along the second row for easier calculation.
|A| = − 0
| −1
−5
−2
0 | | |
+0
3
3
−2
0
− ( − 1)
| |
3
3
−1
−5
= ( − 15 + 3) = − 12
| |
3 −4 5
(ii) Let 1 1 −2
2 3 1
By expanding along the rst row, we have :
Page 5
|A| = 3
| | | | | |
1
3
−2
1
+4
1
2
−2
1
+5
1
2
1
3
= 3(1 + 6) + 4(1 + 4) + 5(3 − 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5 = 46
| |
0 1 2
(iii) Let − 1 0 −3
−2 3 0
By expanding along the rst column, we have :
|A| = 0 | | |
0
3
−3
0
−1
−1
−2
−3
0 | | |
+2
−1
−2
0
3
= − 1( − 6) + 2( − 3)
=6−6=0
| |
2 −1 −2
(iv) Let 0 2 −1
3 −5 0
By expanding along the rst row,we have :
|A| = 2 | −5
2 −1
0 | |
−0
−1
−5
−2
0 | |
+3
−1
2
−2
−1 |
= 2(0 − 5) − 0 + 3(1 + 4)
= − 10 + 15 = 5
Page : 109 , Block Name : Exercise 4.1
[ ]
1 1 −2
Q6 If A = 2 1 − 3 , find | A |
5 4 −9
[ ]
1 1 −2
Answer. Let A = 2 1 − 3 , find | A |
5 4 −9
By expanding along the rst row,we have :
| | | | | |
1
3
−2
1
+4
1
2
−2
1
+5
1
2
1
3
= 3(1 + 6) + 4(1 + 4) + 5(3 − 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5 = 46
| |
0 1 2
(iii) Let − 1 0 −3
−2 3 0
By expanding along the rst column, we have :
|A| = 0 | | |
0
3
−3
0
−1
−1
−2
−3
0 | | |
+2
−1
−2
0
3
= − 1( − 6) + 2( − 3)
=6−6=0
| |
2 −1 −2
(iv) Let 0 2 −1
3 −5 0
By expanding along the rst row,we have :
|A| = 2 | −5
2 −1
0 | |
−0
−1
−5
−2
0 | |
+3
−1
2
−2
−1 |
= 2(0 − 5) − 0 + 3(1 + 4)
= − 10 + 15 = 5
Page : 109 , Block Name : Exercise 4.1
[ ]
1 1 −2
Q6 If A = 2 1 − 3 , find | A |
5 4 −9
[ ]
1 1 −2
Answer. Let A = 2 1 − 3 , find | A |
5 4 −9
By expanding along the rst row,we have :
Page 6
|A| = 1 | | | | | |
1
4
−3
−9
−1
2
5
−3
−9
−2
2
5
1
4
= 1( − 9 + 12) − 1( − 18 + 15) − 2(8 − 5)
= 1(3) − 1( − 3) − 2(3)
=3+3−6
=6−6
=0
Page : 109 , Block Name : Exercise 4.1
Q7 Find values of x, if :
(i) | | | |
2 4
5 1
=
2x
6
4
x
(ii)
| | | |
2 3
4 5
=
x
2x
3
5
Answer.
(i) | | | |
2 4
5 1
=
2x
6
4
x
⇒ 2 × 1 − 5 × 4 = 2x × x − 6 × 4
⇒ 2 − 20 = 2x 2 − 24
⇒ 2x 2 = 6
⇒ x2 = 3
⇒ x = ± √3
(ii) | | | |
2 3
4 5
=
x
2x
3
5
⇒ 2 × 5 − 3 × 4 = x × 5 − 3 × 2x
⇒ 10 − 12 = 5x − 6x
⇒ −2= −x
⇒x=2
Page : 109 , Block Name : Exercise 4.1
Q8 If | | | |
x
18
2
x
=
6
18
2
6
, then x is equal to
(A) 6 (B) = 6 (C) − 6 (D) 0
1
4
−3
−9
−1
2
5
−3
−9
−2
2
5
1
4
= 1( − 9 + 12) − 1( − 18 + 15) − 2(8 − 5)
= 1(3) − 1( − 3) − 2(3)
=3+3−6
=6−6
=0
Page : 109 , Block Name : Exercise 4.1
Q7 Find values of x, if :
(i) | | | |
2 4
5 1
=
2x
6
4
x
(ii)
| | | |
2 3
4 5
=
x
2x
3
5
Answer.
(i) | | | |
2 4
5 1
=
2x
6
4
x
⇒ 2 × 1 − 5 × 4 = 2x × x − 6 × 4
⇒ 2 − 20 = 2x 2 − 24
⇒ 2x 2 = 6
⇒ x2 = 3
⇒ x = ± √3
(ii) | | | |
2 3
4 5
=
x
2x
3
5
⇒ 2 × 5 − 3 × 4 = x × 5 − 3 × 2x
⇒ 10 − 12 = 5x − 6x
⇒ −2= −x
⇒x=2
Page : 109 , Block Name : Exercise 4.1
Q8 If | | | |
x
18
2
x
=
6
18
2
6
, then x is equal to
(A) 6 (B) = 6 (C) − 6 (D) 0
Page 7
Answer. | | | |
18
x 2
x
=
6
18
2
6
⇒ x 2 − 36 = 36 − 36
⇒ x 2 − 36 = 0
⇒ x 2 = 36
⇒x= ±6
Hence, the correct answer is B.
Page : 109 , Block Name : Exercise 4.1
Q1 Using the property of determinants and without expanding, prove that :
| |
x a x+a
y b y+b =0
z c z+c
Answer.
| || || |
x a x+a a x x a x x a a
y b y+b = y b y + y b b =0+0=0
z c z+c z c z z c c
[Here, the two columns of the determinants are identical]
Page : 119 , Block Name : Exercise 4.2
Q2 Using the property of determinants and without expanding,prove that :
| |
a−b b−c c−a
b−c c−a a−b =0
c−a a−b b−c
Answer.
| |
a−b b−c c−a
Δ= b−c c−a a−b
c−a a−b b−c
Applying R 1 → R 1 + R 2, we have :
| |
a−c b−a c−b
Δ= b−c c−a a−b
− (a − c) − (b − a) − (c − b)
18
x 2
x
=
6
18
2
6
⇒ x 2 − 36 = 36 − 36
⇒ x 2 − 36 = 0
⇒ x 2 = 36
⇒x= ±6
Hence, the correct answer is B.
Page : 109 , Block Name : Exercise 4.1
Q1 Using the property of determinants and without expanding, prove that :
| |
x a x+a
y b y+b =0
z c z+c
Answer.
| || || |
x a x+a a x x a x x a a
y b y+b = y b y + y b b =0+0=0
z c z+c z c z z c c
[Here, the two columns of the determinants are identical]
Page : 119 , Block Name : Exercise 4.2
Q2 Using the property of determinants and without expanding,prove that :
| |
a−b b−c c−a
b−c c−a a−b =0
c−a a−b b−c
Answer.
| |
a−b b−c c−a
Δ= b−c c−a a−b
c−a a−b b−c
Applying R 1 → R 1 + R 2, we have :
| |
a−c b−a c−b
Δ= b−c c−a a−b
− (a − c) − (b − a) − (c − b)
Page 8
| |
a−c b−a c−b
= − b−c c−a a−b
a−c b−a c−b
Here, the two rows R 1 and R 3 are identical.
∴Δ=0
Page : 119 , Block Name : Exercise 4.2
Q3 Using the property of determinants and without expanding, prove that :
| |
2 7 65
3 8 75 = 0
5 9 86
Answer.
| | |
2 7 65 2 7 63 + 2
3 8 75 = 3 8 72 + 3
5 9 86 5 9 81 + 5
| || |
2 7 63 2 7 2
= 3 8 72 + 3 8 3
5 9 81 5 9 5
| |
2 7 9(7)
= 3 8 9(8) + 0.......[Two column are identical]
5 9 9(9)
| |
2 7 7
=9 3 8 8 .......[Two column are identical]
5 9 9
=0
Page : 119 , Block Name : Exercise 4.2
Q4 Using the property of determinants and without expanding, prove that :
| |
1 bc a(b + c)
1 ca b(c + a) = 0
1 ab c(a + b)
Answer.
a−c b−a c−b
= − b−c c−a a−b
a−c b−a c−b
Here, the two rows R 1 and R 3 are identical.
∴Δ=0
Page : 119 , Block Name : Exercise 4.2
Q3 Using the property of determinants and without expanding, prove that :
| |
2 7 65
3 8 75 = 0
5 9 86
Answer.
| | |
2 7 65 2 7 63 + 2
3 8 75 = 3 8 72 + 3
5 9 86 5 9 81 + 5
| || |
2 7 63 2 7 2
= 3 8 72 + 3 8 3
5 9 81 5 9 5
| |
2 7 9(7)
= 3 8 9(8) + 0.......[Two column are identical]
5 9 9(9)
| |
2 7 7
=9 3 8 8 .......[Two column are identical]
5 9 9
=0
Page : 119 , Block Name : Exercise 4.2
Q4 Using the property of determinants and without expanding, prove that :
| |
1 bc a(b + c)
1 ca b(c + a) = 0
1 ab c(a + b)
Answer.
Page 9
| |
1 bc a(b + c)
Δ= 1 ca b(c + a)
1 ab c(a + b)
By applying C 3 → C 3 + C 2 , , we have :
| |
1 bc ab + bc + ca
Δ= 1 ca ab + bc + ca
1 ab ab + bc + ca
Here,two columns C1 and C3 are proportional.
∴Δ=0
Page : 119 , Block Name : Exercise 4.2
Q5 Using the property of determinant and without expanding, prove that :
| || |
b+c q+r y+z a p x
c+a r+p z+x =2 b q y
a+b p+q x+y c r z
Answer.
| |
b+c q+r y+z
Δ= c+a r+p z+x
a+b p+q x+y
| || |
b+c q+r y+z b+c q+r y+z
= c+a r+p z+x + c+a r+p z+x
a p x b q y
= Δ 1 + Δ 2( say ) …(i)
| |
b+c q+r y+z
Now, Δ 1 = c + a r+p z+x
a p x
Applying R 2 → R 2 − R 3, we have :
| |
b+c q+r y+z
Δ1 = c r z
a p x
Applying R 1 → R 1 − R 2, we have
1 bc a(b + c)
Δ= 1 ca b(c + a)
1 ab c(a + b)
By applying C 3 → C 3 + C 2 , , we have :
| |
1 bc ab + bc + ca
Δ= 1 ca ab + bc + ca
1 ab ab + bc + ca
Here,two columns C1 and C3 are proportional.
∴Δ=0
Page : 119 , Block Name : Exercise 4.2
Q5 Using the property of determinant and without expanding, prove that :
| || |
b+c q+r y+z a p x
c+a r+p z+x =2 b q y
a+b p+q x+y c r z
Answer.
| |
b+c q+r y+z
Δ= c+a r+p z+x
a+b p+q x+y
| || |
b+c q+r y+z b+c q+r y+z
= c+a r+p z+x + c+a r+p z+x
a p x b q y
= Δ 1 + Δ 2( say ) …(i)
| |
b+c q+r y+z
Now, Δ 1 = c + a r+p z+x
a p x
Applying R 2 → R 2 − R 3, we have :
| |
b+c q+r y+z
Δ1 = c r z
a p x
Applying R 1 → R 1 − R 2, we have
Page 10
| |
b q y
Δ1 = c r z
a p x
Applying R 1 → R 3 and R 2 ↔ R 3 ,we have
| || |
a p x a p x
Δ 1 = ( − 1) 2 b q y = b q y . . . (ii)
c r z c r z
| |
b+c q+r y+z
Δ2 = c + a r+p z+x
b q y
Applying R 2 → R 2 − R 1, we have
| |
c r z
Δ2 = c + a r+p z+x
b q y
Applying R 1 → R 1 − R 3, we have :
| |
c r z
Δ2 = a p x
b q y
Applying R 1 ↦ R 2 and R 2 ↔ R 3, we have
| || |
a p x a p x
Δ 2 = ( − 1) 2 b q y = b q y . . . (iii)
c r z c r z
From, (i),(ii) and (iii), we have :
| |
a p x
Δ=2 b q y
c r z
Hance, the give result is proved.
Page : 119 , Block Name : Exercise 4.2
Q6 By using properties of determinants, show that :
b q y
Δ1 = c r z
a p x
Applying R 1 → R 3 and R 2 ↔ R 3 ,we have
| || |
a p x a p x
Δ 1 = ( − 1) 2 b q y = b q y . . . (ii)
c r z c r z
| |
b+c q+r y+z
Δ2 = c + a r+p z+x
b q y
Applying R 2 → R 2 − R 1, we have
| |
c r z
Δ2 = c + a r+p z+x
b q y
Applying R 1 → R 1 − R 3, we have :
| |
c r z
Δ2 = a p x
b q y
Applying R 1 ↦ R 2 and R 2 ↔ R 3, we have
| || |
a p x a p x
Δ 2 = ( − 1) 2 b q y = b q y . . . (iii)
c r z c r z
From, (i),(ii) and (iii), we have :
| |
a p x
Δ=2 b q y
c r z
Hance, the give result is proved.
Page : 119 , Block Name : Exercise 4.2
Q6 By using properties of determinants, show that :
Page 11
| |
0 a −b
−a 0 −c = 0
b c 0
Answer. We have
| |
0 a −b
Δ = −a 0 −c
b c 0
Applying R 1 → cR 1, we have :
| |
0 ac − bc
1
Δ = c −a 0 −c
b c 0
Applying R 1 → R 1 − bR 2, we have:
| |
ab ac 0
1
Δ = c −a 0 −c
b c 0
| |
b c 0
a
= c −a 0 −c
b c 0
Here, the two rows R1 and R3 are identical.
∴Δ=0
Page : 120 , Block Name : Exercise 4.2
Q7 By using properties of determinants, show that :
| − a2
|
ab ac
ba − b2 bc = 4a 2b 2c 2
ca cb − c2
Answer.
| − a2
|
ab ac
Δ= ba − b2 bc
ca cb − c2
0 a −b
−a 0 −c = 0
b c 0
Answer. We have
| |
0 a −b
Δ = −a 0 −c
b c 0
Applying R 1 → cR 1, we have :
| |
0 ac − bc
1
Δ = c −a 0 −c
b c 0
Applying R 1 → R 1 − bR 2, we have:
| |
ab ac 0
1
Δ = c −a 0 −c
b c 0
| |
b c 0
a
= c −a 0 −c
b c 0
Here, the two rows R1 and R3 are identical.
∴Δ=0
Page : 120 , Block Name : Exercise 4.2
Q7 By using properties of determinants, show that :
| − a2
|
ab ac
ba − b2 bc = 4a 2b 2c 2
ca cb − c2
Answer.
| − a2
|
ab ac
Δ= ba − b2 bc
ca cb − c2
Page 12
| |
−a b c
= abc a −b c ..…..[Taking out factors a,b,c, from R1,R2andR3]
a b −c
| |
−1 1 1
= a 2b 2c 2 1 −1 1 ……[Taking out factors a,b,c from C1,C2and C3]
1 1 −1
Applying R 2 → R 2 + R 1 and R a → R a + R 1, we have :
| |
−1 1 1
Δ = a 2b 2c 2 0 0 2
0 2 0
= a 2b 2c 2( − 1)
| |
0
2
2
0
= − a 2b 2c 2(0 − 4) = 4a 2b 2c 2
Page : 120 , Block Name : Exercise 4.2
Q8 (i) By using properties of determinants, show that :
| | a2
1 a
1 b b 2 = (a − b)(b − c)(c − a)
1 c c2
(ii) By using properties of determinants, show that :
| |
1 1 1
a b c = (a − b)(b − c)(c − a)(a + b + c)
a3 b3 c3
Answer. (i) Let :
| | a2
1 a
Δ= 1 b b2
1 c c2
Applying R 1 → R 1 − R 3 and R 2 → R 2 − R 3, we have :
−a b c
= abc a −b c ..…..[Taking out factors a,b,c, from R1,R2andR3]
a b −c
| |
−1 1 1
= a 2b 2c 2 1 −1 1 ……[Taking out factors a,b,c from C1,C2and C3]
1 1 −1
Applying R 2 → R 2 + R 1 and R a → R a + R 1, we have :
| |
−1 1 1
Δ = a 2b 2c 2 0 0 2
0 2 0
= a 2b 2c 2( − 1)
| |
0
2
2
0
= − a 2b 2c 2(0 − 4) = 4a 2b 2c 2
Page : 120 , Block Name : Exercise 4.2
Q8 (i) By using properties of determinants, show that :
| | a2
1 a
1 b b 2 = (a − b)(b − c)(c − a)
1 c c2
(ii) By using properties of determinants, show that :
| |
1 1 1
a b c = (a − b)(b − c)(c − a)(a + b + c)
a3 b3 c3
Answer. (i) Let :
| | a2
1 a
Δ= 1 b b2
1 c c2
Applying R 1 → R 1 − R 3 and R 2 → R 2 − R 3, we have :
Page 13
| |
0 a−c a2 − c2
Δ= 0 b−c b2 − c2
1 c c2
| |
0 −1 −a − c
= (c − a)(b − c) 0 1 b+c
1 c c2
Applying R 1 → R 1 + R 2, we have
| |
0 0 −a + b
Δ = (b − c)(c − a) 0 1 b+c
1 c c2
| |
0 0 −1
= (a − b)(b − c)(c − a) 0 1 b+c
1 c c2
Expanding along C1 , we have :
Δ = (a − b)(b − c)(c − a) | 0
1
−1
b+c | = (a − b)(b − c)(c − a)
Hence, the given result is proved.
(ii) Let :
| |
1 1 1
Δ= a b c
a3 b3 c3
Applying C 1 → C 1 − C 3 and C 2 → C 2 − C 3
| |
0 0 1
Δ= a−c b−c c
a3 − c3 b3 − c3 c3
|( |
0 0 1
−1 1 c
=(c-a)(b-c)
− a 2 + ac + c 2 ) (b + bc + c ) c
2 2 3
Applying C 1 → C 1 + C 2 , we have
0 a−c a2 − c2
Δ= 0 b−c b2 − c2
1 c c2
| |
0 −1 −a − c
= (c − a)(b − c) 0 1 b+c
1 c c2
Applying R 1 → R 1 + R 2, we have
| |
0 0 −a + b
Δ = (b − c)(c − a) 0 1 b+c
1 c c2
| |
0 0 −1
= (a − b)(b − c)(c − a) 0 1 b+c
1 c c2
Expanding along C1 , we have :
Δ = (a − b)(b − c)(c − a) | 0
1
−1
b+c | = (a − b)(b − c)(c − a)
Hence, the given result is proved.
(ii) Let :
| |
1 1 1
Δ= a b c
a3 b3 c3
Applying C 1 → C 1 − C 3 and C 2 → C 2 − C 3
| |
0 0 1
Δ= a−c b−c c
a3 − c3 b3 − c3 c3
|( |
0 0 1
−1 1 c
=(c-a)(b-c)
− a 2 + ac + c 2 ) (b + bc + c ) c
2 2 3
Applying C 1 → C 1 + C 2 , we have
Page 14
|( ) |
0 0 1
0 1 c
Δ=(c-a)(b-c)
b 2 − a 2 + (bc − ac) (b + bc + c ) c
2 2 3
| ( |
0 0 1
a−c b−c c
=
(a − c) a 2 + ac + c 2 ) (b − c) (b + bc + c ) c
2 2 3
| |
0 0 1
0 1 c
=(b-c)(c-a)(a-b)
− (a + b + c) (b + bc + c ) c
2 2 3
| ( |
0 0 1
0 1 c
=(a-b)(b-c)(c-a)(a+b+c)
−1 b 2 + bc + c 2 ) c 3
Expanding along C1, we have :
Δ = (a − b)(b − c)(c − a)(a + b + c)( − 1)
| | 0
1
1
c
= (a − b)(b − c)(c − a)(a + b + c)
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q9 By using properties of determinants,show that:
| |
x x2 yz
y y2 zx = (x − y)(y − z)(z − x)(xy + yz + zx)
z z2 xy
Answer. Let :
| |
x x2 yz
Δ= y y2 zx
z z2 xy
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1,we have:
0 0 1
0 1 c
Δ=(c-a)(b-c)
b 2 − a 2 + (bc − ac) (b + bc + c ) c
2 2 3
| ( |
0 0 1
a−c b−c c
=
(a − c) a 2 + ac + c 2 ) (b − c) (b + bc + c ) c
2 2 3
| |
0 0 1
0 1 c
=(b-c)(c-a)(a-b)
− (a + b + c) (b + bc + c ) c
2 2 3
| ( |
0 0 1
0 1 c
=(a-b)(b-c)(c-a)(a+b+c)
−1 b 2 + bc + c 2 ) c 3
Expanding along C1, we have :
Δ = (a − b)(b − c)(c − a)(a + b + c)( − 1)
| | 0
1
1
c
= (a − b)(b − c)(c − a)(a + b + c)
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q9 By using properties of determinants,show that:
| |
x x2 yz
y y2 zx = (x − y)(y − z)(z − x)(xy + yz + zx)
z z2 xy
Answer. Let :
| |
x x2 yz
Δ= y y2 zx
z z2 xy
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1,we have:
Page 15
| |
x x2 yz
Δ= y−x y2 − x2 zx − yz
z−x z2 − x2 xy − yz
| |
x x2 yz
= (x − y)(z − x) − 1 −x − y z
1 z+x −y
Applying R 3 → R 3 + R 2,we have :
| |
x x2 yz
Δ = (x − y)(z − x) − 1 −x − y z
0 z−y z−y
| |
x x2 yz
= (x − y)(z − x)(z − y) − 1 −x − y z
0 1 1
Expanding along R3, we have :
Δ = [(x − y)(z − x)(z − y)] ( − 1)
[ | | | x
−1
yz
z
+1
x
−1
x2
−x − y |]
= (x − y)(z − x)(z − y) ( − xz − yz) + [ ( − x − xy + x )]
2 2
= − (x − y)(z − x)(z − y)(xy + yz + zx)
= (x − y)(y − z)(z − x)(xy + yz + zx)
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q10 By using properties of determinants, show that :
| |
x+4 2x 2x
(i) 2x x+4 2x = (5x + 4)(4 − x) 2
2x 2x x+4
| |
y+k y y
(ii) y y+k y = k 2(3y + k)
y y y+k
Answer.
x x2 yz
Δ= y−x y2 − x2 zx − yz
z−x z2 − x2 xy − yz
| |
x x2 yz
= (x − y)(z − x) − 1 −x − y z
1 z+x −y
Applying R 3 → R 3 + R 2,we have :
| |
x x2 yz
Δ = (x − y)(z − x) − 1 −x − y z
0 z−y z−y
| |
x x2 yz
= (x − y)(z − x)(z − y) − 1 −x − y z
0 1 1
Expanding along R3, we have :
Δ = [(x − y)(z − x)(z − y)] ( − 1)
[ | | | x
−1
yz
z
+1
x
−1
x2
−x − y |]
= (x − y)(z − x)(z − y) ( − xz − yz) + [ ( − x − xy + x )]
2 2
= − (x − y)(z − x)(z − y)(xy + yz + zx)
= (x − y)(y − z)(z − x)(xy + yz + zx)
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q10 By using properties of determinants, show that :
| |
x+4 2x 2x
(i) 2x x+4 2x = (5x + 4)(4 − x) 2
2x 2x x+4
| |
y+k y y
(ii) y y+k y = k 2(3y + k)
y y y+k
Answer.
Page 16
| |
x+4 2x 2x
(i) Δ = 2x x+4 2x
2x 2x x+4
Applying R 1 → R 1 + R 2 + R 3,we have :
| |
5x + 4 5x + 4 5x + 4
Δ = 2x x+4 2x
2x 2x x+4
| |
1 1 1
= (5x + 4) 2x x+4 2x
2x 2x x+4
Applying C 2 → C 2 − C 1, C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (5x + 4) 2x −x + 4 0
2x 0 −x + 4
| |
1 1 1
= (5x + 4) 2x x+4 2x
2x 2x x+4
Applying C 2 → C 2 − C 1, C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (5x + 4) 2x −x + 4 0
2x 0 −x + 4
| |
1 0 0
= (5x + 4)(4 − x)(4 − x) 2x 1 0
2x 0 1
Expanding along C3, we have :
Δ = (5x + 4)(4 − x) 2
| |
1
2x
0
1
= (5x + 4)(4 − x) 2
Hence, the given result is proved.
(ii) Let :
x+4 2x 2x
(i) Δ = 2x x+4 2x
2x 2x x+4
Applying R 1 → R 1 + R 2 + R 3,we have :
| |
5x + 4 5x + 4 5x + 4
Δ = 2x x+4 2x
2x 2x x+4
| |
1 1 1
= (5x + 4) 2x x+4 2x
2x 2x x+4
Applying C 2 → C 2 − C 1, C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (5x + 4) 2x −x + 4 0
2x 0 −x + 4
| |
1 1 1
= (5x + 4) 2x x+4 2x
2x 2x x+4
Applying C 2 → C 2 − C 1, C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (5x + 4) 2x −x + 4 0
2x 0 −x + 4
| |
1 0 0
= (5x + 4)(4 − x)(4 − x) 2x 1 0
2x 0 1
Expanding along C3, we have :
Δ = (5x + 4)(4 − x) 2
| |
1
2x
0
1
= (5x + 4)(4 − x) 2
Hence, the given result is proved.
(ii) Let :
Page 17
| |
y+k y y
Δ= y y+k y
y y y+k
Applying R 1 → R 1 + R 2 + R 3,we have :
| |
3y + k 3y + k 3y + k
Δ= y y+k y
y y y+k
| |
1 1 1
= (3y + k) y y+k y
y y y+k
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (3y + k) y k 0
y 0 k
| |
1 0 0
= k 2(3y + k) y 1 0
y 0 1
Expanding along C3 , we have :
Δ = k 2(3y + k) | | 1
y
0
1
= k 2(3y + k)
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q11 By using properties of determinants, show that :
| |
a−b−c 2a 2a
(i) 2b b−c−a 2b = (a + b + c) 3
2c 2c c−a−b
| |
x + y + 2z x y
(ii) z y + z + 2x y = 2(x + y + z) 3
z x z + x + 2y
Answer. (i) Let :
y+k y y
Δ= y y+k y
y y y+k
Applying R 1 → R 1 + R 2 + R 3,we have :
| |
3y + k 3y + k 3y + k
Δ= y y+k y
y y y+k
| |
1 1 1
= (3y + k) y y+k y
y y y+k
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (3y + k) y k 0
y 0 k
| |
1 0 0
= k 2(3y + k) y 1 0
y 0 1
Expanding along C3 , we have :
Δ = k 2(3y + k) | | 1
y
0
1
= k 2(3y + k)
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q11 By using properties of determinants, show that :
| |
a−b−c 2a 2a
(i) 2b b−c−a 2b = (a + b + c) 3
2c 2c c−a−b
| |
x + y + 2z x y
(ii) z y + z + 2x y = 2(x + y + z) 3
z x z + x + 2y
Answer. (i) Let :
Page 18
| |
a−b−c 2a 2a
Δ= 2b b−c−a 2b
2c 2c c−a−b
Applying R 1 → R 1 + R 2 + R 3, we have :
| |
a+b+c a+b+c a+b+c
Δ = 2b b−c−a 2b
2c 2c c−a−b
| |
1 1 1
= (a + b + c) 2b b−c−a 2b
2c 2c c−a−b
Applying C 2 → C 2 − C 1, C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (a + b + c) 2b − (a + b + c) 0
2c 0 − (a + b + c)
| |
1 0 0
= (a + b + c) 3 2b −1 0
2c 0 −1
Expanding along C3, we have :
Δ = (a + b + c) 3( − 1)( − 1) = (a + b + c) 3
Hence, the given result is proved.
| |
x + y + 2z x y
(ii) Δ = z y + z + 2x y
z x z + x + 2y
Applying C 1 → C 1 + C 2 + C 3, we have :
| |
2(x + y + z) x y
Δ = 2(x + y + z) y + z + 2x y
2(x + y + z) x z + x + 2y
| |
1 x y
= 2(x + y + z) 1 y + z + 2x y
1 x z + x + 2y
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
a−b−c 2a 2a
Δ= 2b b−c−a 2b
2c 2c c−a−b
Applying R 1 → R 1 + R 2 + R 3, we have :
| |
a+b+c a+b+c a+b+c
Δ = 2b b−c−a 2b
2c 2c c−a−b
| |
1 1 1
= (a + b + c) 2b b−c−a 2b
2c 2c c−a−b
Applying C 2 → C 2 − C 1, C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = (a + b + c) 2b − (a + b + c) 0
2c 0 − (a + b + c)
| |
1 0 0
= (a + b + c) 3 2b −1 0
2c 0 −1
Expanding along C3, we have :
Δ = (a + b + c) 3( − 1)( − 1) = (a + b + c) 3
Hence, the given result is proved.
| |
x + y + 2z x y
(ii) Δ = z y + z + 2x y
z x z + x + 2y
Applying C 1 → C 1 + C 2 + C 3, we have :
| |
2(x + y + z) x y
Δ = 2(x + y + z) y + z + 2x y
2(x + y + z) x z + x + 2y
| |
1 x y
= 2(x + y + z) 1 y + z + 2x y
1 x z + x + 2y
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
Page 19
| |
1 x y
Δ = 2(x + y + z) 0 x+y+z 0
0 0 x+y+z
| |
1 x y
= 2(x + y + z) 3 0 1 0
0 0 1
Expanding along R3, we have :
Δ = 2(x + y + z) 3(1)(1 − 0) = 2(x + y + z) 3
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q12 By Using properties of determinants,show that :
| | x2
1 x
x2 1 x (
= 1 − x3 ) 2
x x2 1
Answer.
| | x2
1 x
Δ = x2 1 x
x x2 1
Applying R 1 → R 1 + R 2 + R 3, we have :
| |
1 + x + x2 1 + x + x2 1 + x + x2
Δ= x2 1 x
x x2 1
| |
1 1 1
x2 1 x
(
= 1+x+x 2
)x 2 1 x
x x2 1
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
1 x y
Δ = 2(x + y + z) 0 x+y+z 0
0 0 x+y+z
| |
1 x y
= 2(x + y + z) 3 0 1 0
0 0 1
Expanding along R3, we have :
Δ = 2(x + y + z) 3(1)(1 − 0) = 2(x + y + z) 3
Hence, the given result is proved.
Page : 120 , Block Name : Exercise 4.2
Q12 By Using properties of determinants,show that :
| | x2
1 x
x2 1 x (
= 1 − x3 ) 2
x x2 1
Answer.
| | x2
1 x
Δ = x2 1 x
x x2 1
Applying R 1 → R 1 + R 2 + R 3, we have :
| |
1 + x + x2 1 + x + x2 1 + x + x2
Δ= x2 1 x
x x2 1
| |
1 1 1
x2 1 x
(
= 1+x+x 2
)x 2 1 x
x x2 1
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
Page 20
| |
1 0 0
(
Δ = 1 + x + x2 ) x 1−x x−x
2 2 2
x x2 − x 1−x
| |
1 0 0
(
= 1 + x + x 2 (1 − x)(1 − x)) x2 1+x x
x −x 1
| |
1 0 0
( )
= 1 − x 3 (1 − x) x 2 1+x x
x −x 1
Expanding along R 1, we have:
(
Δ = 1 − x 3 (1 − x)(1) ) | 1+x
−x
x
1 |
( ) (
= 1 − x 3 (1 − x) 1 + x + x 2 )
= (1 − x )(1 − x )
3 3
= (1 − x ) 3 2
Hence, the given result is proved.
Page : 121 , Block Name : Exercise 4.2
Q13 By using properties of determinants, show that :
| |
1 + a2 − b2 2ab − 2b
2ab 1 − a2 + b2 2a (
= 1 + a2 + b2 ) 3
2b − 2a 1 − a2 − b2
Answer.
| |
1 + a2 − b2 2ab − 2b
Δ= 2ab 1 − a2 + b2 2a
2b − 2a 1 − a2 − b2
Applying R 1 → R 1 + bR 3 and R 2 → R 2 − aR 3,we have :
1 0 0
(
Δ = 1 + x + x2 ) x 1−x x−x
2 2 2
x x2 − x 1−x
| |
1 0 0
(
= 1 + x + x 2 (1 − x)(1 − x)) x2 1+x x
x −x 1
| |
1 0 0
( )
= 1 − x 3 (1 − x) x 2 1+x x
x −x 1
Expanding along R 1, we have:
(
Δ = 1 − x 3 (1 − x)(1) ) | 1+x
−x
x
1 |
( ) (
= 1 − x 3 (1 − x) 1 + x + x 2 )
= (1 − x )(1 − x )
3 3
= (1 − x ) 3 2
Hence, the given result is proved.
Page : 121 , Block Name : Exercise 4.2
Q13 By using properties of determinants, show that :
| |
1 + a2 − b2 2ab − 2b
2ab 1 − a2 + b2 2a (
= 1 + a2 + b2 ) 3
2b − 2a 1 − a2 − b2
Answer.
| |
1 + a2 − b2 2ab − 2b
Δ= 2ab 1 − a2 + b2 2a
2b − 2a 1 − a2 − b2
Applying R 1 → R 1 + bR 3 and R 2 → R 2 − aR 3,we have :
Page 21
| |
1 + a2 + b2 0 ( )
− b 1 + a2 + b2
a (1 + a + b )
Δ=
0 1 + a2 + b2 2 2
2b − 2a 1 − a2 − b2
| |
1 0 −b
= (1 + a2 + b2
2
) 0 1 a
2b − 2a 1 − a2 − b2
Expanding along R 1, we have:
(
Δ = 1 + a2 + b2 ) 2
[|
(1)
− 2a
1 a
1 − a2 − b2
− b
0
2b| | 1
− 2a |
(
= 1 + a2 + b2) [1 − a − b + 2a − b( − 2b) ]
2 2 2 2
= (1 + a + b ) (1 + a + b )
2 2 2 2 2
= (1 + a + b )2 2 3
Page : 121 , Block Name : Exercise 4.2
Q14 By using properties of determinants,show that :
| |
a2 + 1 ab ac
ab b2 + 1 bc = 1 + a2 + b2 + c2
ca cb c2 + 1
Answer.
| |
a2 + 1 ab ac
Δ= ab b2 + 1 bc
ca cb c2 + 1
Taking out common factors a,b and c from R1,R2 and R3 respectively, we have :
1 + a2 + b2 0 ( )
− b 1 + a2 + b2
a (1 + a + b )
Δ=
0 1 + a2 + b2 2 2
2b − 2a 1 − a2 − b2
| |
1 0 −b
= (1 + a2 + b2
2
) 0 1 a
2b − 2a 1 − a2 − b2
Expanding along R 1, we have:
(
Δ = 1 + a2 + b2 ) 2
[|
(1)
− 2a
1 a
1 − a2 − b2
− b
0
2b| | 1
− 2a |
(
= 1 + a2 + b2) [1 − a − b + 2a − b( − 2b) ]
2 2 2 2
= (1 + a + b ) (1 + a + b )
2 2 2 2 2
= (1 + a + b )2 2 3
Page : 121 , Block Name : Exercise 4.2
Q14 By using properties of determinants,show that :
| |
a2 + 1 ab ac
ab b2 + 1 bc = 1 + a2 + b2 + c2
ca cb c2 + 1
Answer.
| |
a2 + 1 ab ac
Δ= ab b2 + 1 bc
ca cb c2 + 1
Taking out common factors a,b and c from R1,R2 and R3 respectively, we have :
Page 22
| |
1
a+ a b c
1
Δ = abc a b+ b c
1
a b c+ c
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have :
| |
1
a+ a b c
1 1
Δ = abc −a b
0
1 1
−a 0 c
Applying C 1 → ac 1, C 2 → bC 2 , and C → cC ,we have :
3 3
| |
a2 + 1 b2 c2
1
Δ = abc × abc − 1 1 0
−1 0 1
| |
a2 + 1 b2 c2
= −1 1 0
−1 0 1
Expanding along R3, we have :
Δ= −1
| | |
b2
1
c2
0
+1
a2 + 1
−1
b2
1 |
( ) (
= − 1 − c2 + a2 + 1 + b2 = 1 + a2 + b2 + c2 )
Hence, the given result is proved.
Page : 121 , Block Name : Exercise 4.2
Q15 Let A be a square matrix of order 3 × 3, then | kA | is equal
to
(A) k | A | (B) k 2 | A | (C) k 3 | A | (D) 3k | A |
Answer. Answer is C :
A is a Square matrix of order 3 × 3.
1
a+ a b c
1
Δ = abc a b+ b c
1
a b c+ c
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have :
| |
1
a+ a b c
1 1
Δ = abc −a b
0
1 1
−a 0 c
Applying C 1 → ac 1, C 2 → bC 2 , and C → cC ,we have :
3 3
| |
a2 + 1 b2 c2
1
Δ = abc × abc − 1 1 0
−1 0 1
| |
a2 + 1 b2 c2
= −1 1 0
−1 0 1
Expanding along R3, we have :
Δ= −1
| | |
b2
1
c2
0
+1
a2 + 1
−1
b2
1 |
( ) (
= − 1 − c2 + a2 + 1 + b2 = 1 + a2 + b2 + c2 )
Hence, the given result is proved.
Page : 121 , Block Name : Exercise 4.2
Q15 Let A be a square matrix of order 3 × 3, then | kA | is equal
to
(A) k | A | (B) k 2 | A | (C) k 3 | A | (D) 3k | A |
Answer. Answer is C :
A is a Square matrix of order 3 × 3.
Page 23
[ ]
a1 b1 c1
Let A = a2 b2 c2
a3 b3 c3
[ ]
a1 b1 c1
A= a2 b2 c2
a3 b3 c3
[ ]
ka 1 kb 1 kc 1
Then, kA = ka 2 kb 2 kc 2
ka 3 kb 3 kc 3
| |
ka 1 kb 1 kc 1
∴ | kA | = ka 2 kb 2 kc 2
ka 3 kb 3 kc 3
| |
a1 b1 c1
= k3 a2 b2 c 2 …..[Taking out common factors k from each row]
a3 b3 c3
= k3 | A |
∴ | kA | = k 3 | A |
Hence, the correct answer is C.
Page : 121 , Block Name : Exercise 4.2
Q16 Which of the following is correct
Determinant is a square matrix.
Determinant is a number associated to a matrix.
Determinant is a number associated to a square matrix.
None of these
Answer. We know that to every square matrix, A=[aij] of order n.
We can associate a number called the determinant of square matrix A,
where aij=(i,j)th element of A.
Hence, the correct answer is C.
Page : 121 , Block Name : Exercise 4.2
a1 b1 c1
Let A = a2 b2 c2
a3 b3 c3
[ ]
a1 b1 c1
A= a2 b2 c2
a3 b3 c3
[ ]
ka 1 kb 1 kc 1
Then, kA = ka 2 kb 2 kc 2
ka 3 kb 3 kc 3
| |
ka 1 kb 1 kc 1
∴ | kA | = ka 2 kb 2 kc 2
ka 3 kb 3 kc 3
| |
a1 b1 c1
= k3 a2 b2 c 2 …..[Taking out common factors k from each row]
a3 b3 c3
= k3 | A |
∴ | kA | = k 3 | A |
Hence, the correct answer is C.
Page : 121 , Block Name : Exercise 4.2
Q16 Which of the following is correct
Determinant is a square matrix.
Determinant is a number associated to a matrix.
Determinant is a number associated to a square matrix.
None of these
Answer. We know that to every square matrix, A=[aij] of order n.
We can associate a number called the determinant of square matrix A,
where aij=(i,j)th element of A.
Hence, the correct answer is C.
Page : 121 , Block Name : Exercise 4.2
Page 24
Q1 Find area of the triangle with vertices at the point given in each of the following :
(i) (1,0),(6,0),(4,3)
(ii) (2,7,(1,1),(10,8)
(iii) (-2,-3),(3,2),(-1,-8)
Answer. (i) the are of the triangle with vertices (1,0),(6,0),(4,3) is given by the relation,
| |
1 0 1
1
Δ= 2 6 0 1
4 3 1
1
= [1(0 − 3) − 0(6 − 4) + 1(18 − 0)]
2
1 15
= [ − 3 + 18] = square units
2 2
(ii) The area of the triangle with vertices (2,7),(1,1),(10,8) is given by the relation,
| |
2 7 1
1
Δ= 1 1 1
2
10 8 1
1
= [2(1 − 8) − 7(1 − 10) + 1(8 − 10)]
2
1 1
= [ − 14 + 63 − 2] = [ − 16 + 63]
2 2
47
= square units
2
(iii) The area of the triangle with vertices (-2,-3),(3,2),(-1,-8) is given by the relation,
| |
−2 −3 1
1
Δ= 3 2 1
2
−1 −8 1
1
= [ − 2(2 + 8) + 3(3 + 1) + 1( − 24 + 2)]
2
1
= [ − 2(10) + 3(4) + 1( − 22)]
2
1
= [ − 20 + 12 − 22]
2
30
= − = − 15
2
Hence, the area of triangle is | − 15 | = 15 square units.
Page : 122 , Block Name : Exercise 4.3
(i) (1,0),(6,0),(4,3)
(ii) (2,7,(1,1),(10,8)
(iii) (-2,-3),(3,2),(-1,-8)
Answer. (i) the are of the triangle with vertices (1,0),(6,0),(4,3) is given by the relation,
| |
1 0 1
1
Δ= 2 6 0 1
4 3 1
1
= [1(0 − 3) − 0(6 − 4) + 1(18 − 0)]
2
1 15
= [ − 3 + 18] = square units
2 2
(ii) The area of the triangle with vertices (2,7),(1,1),(10,8) is given by the relation,
| |
2 7 1
1
Δ= 1 1 1
2
10 8 1
1
= [2(1 − 8) − 7(1 − 10) + 1(8 − 10)]
2
1 1
= [ − 14 + 63 − 2] = [ − 16 + 63]
2 2
47
= square units
2
(iii) The area of the triangle with vertices (-2,-3),(3,2),(-1,-8) is given by the relation,
| |
−2 −3 1
1
Δ= 3 2 1
2
−1 −8 1
1
= [ − 2(2 + 8) + 3(3 + 1) + 1( − 24 + 2)]
2
1
= [ − 2(10) + 3(4) + 1( − 22)]
2
1
= [ − 20 + 12 − 22]
2
30
= − = − 15
2
Hence, the area of triangle is | − 15 | = 15 square units.
Page : 122 , Block Name : Exercise 4.3
Page 25
Q2 Show that points
A(a, b + c), B(b, c + a), C(c, a + b) are collinear.
Answer. Area of Δ ABC is given by the relation,
| |
a b+c 1
1
Δ= 2 b c+a 1
c a+b 1
| |
a b+c 1
1
= 2 b−a a−b 0 (Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1)
c−a a−c 0
| |
a b+c 1
1
= 2 (a − b)(c − a) − 1 1 0
1 −1 0
| |(
a b+c 1
1
= 2 (a − b)(c − a) − 1 1 0 Applying R 3 → R 3 + R 2 )
0 0 0
= 0 (All elements of R3 are 0)
Thus, the area of the triangle formed by point A,B, and c is zero.
Hence, the points A, B, and C are collinear.
Page : 123 , Block Name : Exercise 4.3
Q3 Find values of k if area of triangle is 4sq. units and vertices are
(i) (k, 0),(4,0),(0,2)
(ii) (-2,0),(0,4),(0,k)
Answer. We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and
(x3, y3) is the absolute value of the determinant (Δ), where
| |
x1 y1 1
1
Δ = 2 x2 y2 1
x3 y3 1
It is given that the area of triangle is 4 square units.
∴Δ= ±4
(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
A(a, b + c), B(b, c + a), C(c, a + b) are collinear.
Answer. Area of Δ ABC is given by the relation,
| |
a b+c 1
1
Δ= 2 b c+a 1
c a+b 1
| |
a b+c 1
1
= 2 b−a a−b 0 (Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1)
c−a a−c 0
| |
a b+c 1
1
= 2 (a − b)(c − a) − 1 1 0
1 −1 0
| |(
a b+c 1
1
= 2 (a − b)(c − a) − 1 1 0 Applying R 3 → R 3 + R 2 )
0 0 0
= 0 (All elements of R3 are 0)
Thus, the area of the triangle formed by point A,B, and c is zero.
Hence, the points A, B, and C are collinear.
Page : 123 , Block Name : Exercise 4.3
Q3 Find values of k if area of triangle is 4sq. units and vertices are
(i) (k, 0),(4,0),(0,2)
(ii) (-2,0),(0,4),(0,k)
Answer. We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and
(x3, y3) is the absolute value of the determinant (Δ), where
| |
x1 y1 1
1
Δ = 2 x2 y2 1
x3 y3 1
It is given that the area of triangle is 4 square units.
∴Δ= ±4
(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
Page 26
| |
k 0 1
1
Δ= 2 4 0 1
0 2 1
1
= 2 [k(0 − 2) − 0(4 − 0) + 1(8 − 0)]
1
= 2 [ − 2k + 8] = − k + 4
∴ −k+4= ±4
When − k + 4 = − 4, k = 8
When − k + 4 = 4, k = 0
Hence, k = 0, 8
(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,
| |
−2 0 1
1
Δ= 2 0 4 1
0 k 1
1
= [ − 2(4 − k)]
2
=k−4
∴k−4= ±4
When k − 4 = − 4, k = 0
When k − 4 = 4, k = 8
Hence, k = 0, 8
Page : 123 , Block Name : Exercise 4.3
Q4 (i) Find equation of line joining (1, 2) and (3, 6) using determinants
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants
Answer. (i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6).
Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
| |
1 2 1
1
∴ 3 6 1 =0
2
x y 1
1
⇒ [1(6 − y) − 2(3 − x) + 1(3y − 6x)] = 0
2
⇒ 6 − y − 6 + 2x + 3y − 6x = 0
⇒ 2y − 4x = 0
⇒ y = 2x
Hence, the equation of the line joining the given points is y = 2x.
k 0 1
1
Δ= 2 4 0 1
0 2 1
1
= 2 [k(0 − 2) − 0(4 − 0) + 1(8 − 0)]
1
= 2 [ − 2k + 8] = − k + 4
∴ −k+4= ±4
When − k + 4 = − 4, k = 8
When − k + 4 = 4, k = 0
Hence, k = 0, 8
(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,
| |
−2 0 1
1
Δ= 2 0 4 1
0 k 1
1
= [ − 2(4 − k)]
2
=k−4
∴k−4= ±4
When k − 4 = − 4, k = 0
When k − 4 = 4, k = 8
Hence, k = 0, 8
Page : 123 , Block Name : Exercise 4.3
Q4 (i) Find equation of line joining (1, 2) and (3, 6) using determinants
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants
Answer. (i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6).
Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
| |
1 2 1
1
∴ 3 6 1 =0
2
x y 1
1
⇒ [1(6 − y) − 2(3 − x) + 1(3y − 6x)] = 0
2
⇒ 6 − y − 6 + 2x + 3y − 6x = 0
⇒ 2y − 4x = 0
⇒ y = 2x
Hence, the equation of the line joining the given points is y = 2x.
Page 27
(ii) Let P (x, y) be any point on the line joining points A (3, 1) and
B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
| |
3 1 1
1
∴ 2 9 3 1 =0
x y 1
1
⇒ 2 [3(3 − y) − 1(9 − x) + 1(9y − 3x)] = 0
⇒ 9 − 3y − 9 + x + 9y − 3x = 0
⇒ 6y − 2x = 0
⇒ x − 3y = 0
Hence, the equation of the line joining the given points is x − 3y = 0
Page : 123 , Block Name : Exercise 4.3
Q5 If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2
Answer. (D),
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
| |
2 −6 1
1
Δ= 5 4 1
2
k 4 1
1
= [2(4 − 4) + 6(5 − k) + 1(20 − 4k)]
2
1
= [30 − 6k + 20 − 4k]
2
1
= [50 − 10k]
2
= 25 − 5k
It is given that the area of the triangle is ±35.
Therefore, we have:
⇒ 25 − 5k = ± 35
⇒ 5(5 − k) = ± 35
⇒5−k= ±7
When 5 − k = − 7, k = 5 + 7 = 12
When 5 − k = 7, k = 5 − 7 = − 2
Hence, k = 12, − 2
So The correct answer is D.
Page : 123 , Block Name : Exercise 4.3
B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
| |
3 1 1
1
∴ 2 9 3 1 =0
x y 1
1
⇒ 2 [3(3 − y) − 1(9 − x) + 1(9y − 3x)] = 0
⇒ 9 − 3y − 9 + x + 9y − 3x = 0
⇒ 6y − 2x = 0
⇒ x − 3y = 0
Hence, the equation of the line joining the given points is x − 3y = 0
Page : 123 , Block Name : Exercise 4.3
Q5 If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2
Answer. (D),
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
| |
2 −6 1
1
Δ= 5 4 1
2
k 4 1
1
= [2(4 − 4) + 6(5 − k) + 1(20 − 4k)]
2
1
= [30 − 6k + 20 − 4k]
2
1
= [50 − 10k]
2
= 25 − 5k
It is given that the area of the triangle is ±35.
Therefore, we have:
⇒ 25 − 5k = ± 35
⇒ 5(5 − k) = ± 35
⇒5−k= ±7
When 5 − k = − 7, k = 5 + 7 = 12
When 5 − k = 7, k = 5 − 7 = − 2
Hence, k = 12, − 2
So The correct answer is D.
Page : 123 , Block Name : Exercise 4.3
Page 28
Q1 Write Minors and Cofactors of the elements of following determinants:
(i) | | | |
2
0
−4
3
(ii)
a
b
c
d
Answer.(i)The given determinant is | | 2
0
−4
3
Minor of element aij is Mij.
∴M11 = minor of element a11 = 3
M12 = minor of element a12 = 0
M21 = minor of element a21 = −4
M22 = minor of element a22 = 2
Cofactor of aij is Aij = (−1)i+j ij.
∴ A 11 = ( − 1) 1 + 1M 11 = ( − 1) 2(3) = 3
A 12 = ( − 1) 1 + 2M 12 = ( − 1) 3(0) = 0
A 21 = ( − 1) 2 + 1M 21 = ( − 1) 3( − 4) = 4
A 22 = ( − 1) 2 + 2M 22 = ( − 1) 4(2) = 2
(ii) The given determinant is
| |
a
b
c
d
.
Minor of element a ij is M ij
∴ M 11 = minor of element a 11 = d
M 12 = minor of element a 12 = b
M 21 = minor of element a 21 = c
M 22 = minor of element a 22 = a
cofactor of a ij is A ij = ( − 1) i + jM ij
∴ A 11 = ( − 1) 1 + 1M 11 = ( − 1) 2(d) = d
A 12 = ( − 1) 1 + 2M 12 = ( − 1) 3(b) = − b
A 21 = ( − 1) 2 + 1M 21 = ( − 1) 3(c) = − b
A 22 = ( − 1) 2 + 2M 22 = ( − 1) 4(a) = a
Page : 126 , Block Name : Exercise 4.4
(i) | | | |
2
0
−4
3
(ii)
a
b
c
d
Answer.(i)The given determinant is | | 2
0
−4
3
Minor of element aij is Mij.
∴M11 = minor of element a11 = 3
M12 = minor of element a12 = 0
M21 = minor of element a21 = −4
M22 = minor of element a22 = 2
Cofactor of aij is Aij = (−1)i+j ij.
∴ A 11 = ( − 1) 1 + 1M 11 = ( − 1) 2(3) = 3
A 12 = ( − 1) 1 + 2M 12 = ( − 1) 3(0) = 0
A 21 = ( − 1) 2 + 1M 21 = ( − 1) 3( − 4) = 4
A 22 = ( − 1) 2 + 2M 22 = ( − 1) 4(2) = 2
(ii) The given determinant is
| |
a
b
c
d
.
Minor of element a ij is M ij
∴ M 11 = minor of element a 11 = d
M 12 = minor of element a 12 = b
M 21 = minor of element a 21 = c
M 22 = minor of element a 22 = a
cofactor of a ij is A ij = ( − 1) i + jM ij
∴ A 11 = ( − 1) 1 + 1M 11 = ( − 1) 2(d) = d
A 12 = ( − 1) 1 + 2M 12 = ( − 1) 3(b) = − b
A 21 = ( − 1) 2 + 1M 21 = ( − 1) 3(c) = − b
A 22 = ( − 1) 2 + 2M 22 = ( − 1) 4(a) = a
Page : 126 , Block Name : Exercise 4.4
Page 29
| | | |
1 0 0 1 0 4
Q2 (i) 0 1 0 (ii) 3 5 −1
0 0 1 0 1 2
| |
1 0 0
Answer. (i) The given determinant is 0 1 0 .
0 0 1
By the de nition of minors and cofactors, we have:
M 11 = minor of a 11 = | |
1 0
0 1
=1
M 12 = minor of a 13 = | |
0
0
0
1
=0
M 13 = minor of a 13 = | |
0
0
1
0
=0
M 21 = minor of a 21 = | |
0
0
0
1
=0
M 22 = minor of a 22 = | |
1
0
0
1
=1
M 23 = minor of a 23 = | |
1
0
0
0
=0
M 31 = minor of a 31 = | |
0
1
0
0
=0
M 32 = minor of a 32 = | |
1
0
0
0
=0
M 33 = minor of a 33 = | |
1
0
0
1
=1
A 11 = cofactor of a 11 = ( − 1) 1 + 1M 11 = 1
A 12 = cofactor of a 12 = ( − 1) 1 + 2M 12 = 0
A 13 = cofactor of a 13 = ( − 1) 1 + 3M 13 = 0
A 21 = cofactor of a 21 = ( − 1) 2 + 1M 21 = 0
A 22 = cofactor of a 22 = ( − 1) 2 + 2M 22 = 1
A 23 = cofactor of a 23 = ( − 1) 2 + 3M 23 = 0
1 0 0 1 0 4
Q2 (i) 0 1 0 (ii) 3 5 −1
0 0 1 0 1 2
| |
1 0 0
Answer. (i) The given determinant is 0 1 0 .
0 0 1
By the de nition of minors and cofactors, we have:
M 11 = minor of a 11 = | |
1 0
0 1
=1
M 12 = minor of a 13 = | |
0
0
0
1
=0
M 13 = minor of a 13 = | |
0
0
1
0
=0
M 21 = minor of a 21 = | |
0
0
0
1
=0
M 22 = minor of a 22 = | |
1
0
0
1
=1
M 23 = minor of a 23 = | |
1
0
0
0
=0
M 31 = minor of a 31 = | |
0
1
0
0
=0
M 32 = minor of a 32 = | |
1
0
0
0
=0
M 33 = minor of a 33 = | |
1
0
0
1
=1
A 11 = cofactor of a 11 = ( − 1) 1 + 1M 11 = 1
A 12 = cofactor of a 12 = ( − 1) 1 + 2M 12 = 0
A 13 = cofactor of a 13 = ( − 1) 1 + 3M 13 = 0
A 21 = cofactor of a 21 = ( − 1) 2 + 1M 21 = 0
A 22 = cofactor of a 22 = ( − 1) 2 + 2M 22 = 1
A 23 = cofactor of a 23 = ( − 1) 2 + 3M 23 = 0
Page 30
A 31 = cofactor of a 31 = ( − 1) 3 + 1M 31 = 0
A 32 = cofactor of a 32 = ( − 1) 3 + 2M 32 = 0
A 33 = cofactor of a 33 = ( − 1) 3 + 3M 33 = 1
| |
1 0 4
(ii) The given determinant is 3 5 −1
0 1 2
By de nition of minors and cofactors, we have:
M 11 = minor of a 11 = | |
5
1
−1
2
= 10 + 1 = 11
M 12 = minor of a 12 = | |
3
0
−1
2
=6−0=6
M 13 = minor of a 13 = | |
3
0
5
1
=3−0=3
M 21 = minor of a 21 = | |
0
1
4
2
=0−4= −4
M 22 = minor of a 22 = | |
1 4
0 2
=2−0=2
M 23 = minor of a 23 = | |
1
0
0
1
=1−0=1
M 31 = minor of a 31 = | |
0
5 −1
4
= 0 − 20 = − 20
M 32 = minor of a 32 =
1
3| | −1
4
= − 1 − 12 = − 13
M 33 = minor of a 33 = | |
1
3
0
5
=5−0=5
A 11 = cofactor of a 11 = ( − 1) 1 + 1M 11 = 11
A 12 = cofactor of a 12 = ( − 1) 1 + 2M 12 = − 6
A 13 = cofactor of a 13 = ( − 1) 1 + 3M 13 = 3
A 21 = cofactor of a 21 = ( − 1) 2 + 1M 21 = 4
A 22 = cofactor of a 22 = ( − 1) 2 + 2M 22 = 2
A 23 = cofactor of a 23 = ( − 1) 2 + 3M 23 = − 1
A 32 = cofactor of a 32 = ( − 1) 3 + 2M 32 = 0
A 33 = cofactor of a 33 = ( − 1) 3 + 3M 33 = 1
| |
1 0 4
(ii) The given determinant is 3 5 −1
0 1 2
By de nition of minors and cofactors, we have:
M 11 = minor of a 11 = | |
5
1
−1
2
= 10 + 1 = 11
M 12 = minor of a 12 = | |
3
0
−1
2
=6−0=6
M 13 = minor of a 13 = | |
3
0
5
1
=3−0=3
M 21 = minor of a 21 = | |
0
1
4
2
=0−4= −4
M 22 = minor of a 22 = | |
1 4
0 2
=2−0=2
M 23 = minor of a 23 = | |
1
0
0
1
=1−0=1
M 31 = minor of a 31 = | |
0
5 −1
4
= 0 − 20 = − 20
M 32 = minor of a 32 =
1
3| | −1
4
= − 1 − 12 = − 13
M 33 = minor of a 33 = | |
1
3
0
5
=5−0=5
A 11 = cofactor of a 11 = ( − 1) 1 + 1M 11 = 11
A 12 = cofactor of a 12 = ( − 1) 1 + 2M 12 = − 6
A 13 = cofactor of a 13 = ( − 1) 1 + 3M 13 = 3
A 21 = cofactor of a 21 = ( − 1) 2 + 1M 21 = 4
A 22 = cofactor of a 22 = ( − 1) 2 + 2M 22 = 2
A 23 = cofactor of a 23 = ( − 1) 2 + 3M 23 = − 1
Page 31
A 31 = cofactor of a 31 = ( − 1) 3 + 1M 31 = − 20
A 32 = cofactor of a 32 = ( − 1) 3 + 2M 32 = 13
A 33 = cofactor of a 33 = ( − 1) 3 + 3M 33 = 5
Page : 126 , Block Name : Exercise 4.4
| |
5 3 8
Q3 Using Cofactors of elements of second row, evaluate Δ = 2 0 1
1 2 3
| |
5 3 8
Answer. The given determinant is 2 0 1
1 2 3
We have:
| |
3
2
8
3
= 9 − 16 = − 7
∴ A 21 = cofactor of a 21 = ( − 1) 2 + 1M 21 = 7
M 23 =
| |
5 3
1 2
= 10 − 3 = 7
We know that Δ is equal to the sum of the product of the elements of the second row with their
corresponding cofactors.
∴ Δ = a 21A 21 + a 22A 22 + a 23A 23 = 2(7) + 0(7) + 1( − 7) = 14 − 7 = 7
Page : 126 , Block Name : Exercise 4.4
Q4 Using Cofactors of elements of third column, evaluate
| |
1 x yz
Δ= 1 y zx
1 z xy
| |
1 x yz
Answer. The given determinant is 1 y zx
1 z xy
We have:
M 13 =
| |
1
1
y
z
=z−y
A 32 = cofactor of a 32 = ( − 1) 3 + 2M 32 = 13
A 33 = cofactor of a 33 = ( − 1) 3 + 3M 33 = 5
Page : 126 , Block Name : Exercise 4.4
| |
5 3 8
Q3 Using Cofactors of elements of second row, evaluate Δ = 2 0 1
1 2 3
| |
5 3 8
Answer. The given determinant is 2 0 1
1 2 3
We have:
| |
3
2
8
3
= 9 − 16 = − 7
∴ A 21 = cofactor of a 21 = ( − 1) 2 + 1M 21 = 7
M 23 =
| |
5 3
1 2
= 10 − 3 = 7
We know that Δ is equal to the sum of the product of the elements of the second row with their
corresponding cofactors.
∴ Δ = a 21A 21 + a 22A 22 + a 23A 23 = 2(7) + 0(7) + 1( − 7) = 14 − 7 = 7
Page : 126 , Block Name : Exercise 4.4
Q4 Using Cofactors of elements of third column, evaluate
| |
1 x yz
Δ= 1 y zx
1 z xy
| |
1 x yz
Answer. The given determinant is 1 y zx
1 z xy
We have:
M 13 =
| |
1
1
y
z
=z−y
Page 32
M 23 = | | 1
1
x
z
=z−x
M 33 = | | 1 x
1 y
=y−x
∴ A 13 = cofactor of a 13 = ( − 1) 1 + 3M 13 = (z − y)
A 23 = cofactor of a 23 = ( − 1) 2 + 3M 23 = − (z − x) = (x − z)
A 33 = cofactor of a 33 = ( − 1) 3 + 3M 33 = (y − x)
We know that Δ is equal to the sum of the product of the elements of the second row with their
corresponding cofactors.
∴ Δ = a 13A 13 + a 23A 23 + a 33A 33
= yz(z − y) + zx(x − z) + xy(y − x)
= yz 2 − y 2z + x 2z − xz 2 + xy 2 − x 2y
( ) ( ) (
= x 2z − y 2z + yz 2 − xz 2 + xy 2 − x 2y )
= z (x − y ) + z (y − x) + xy(y − x)
2 2
= z(x − y)(x + y) + z 2(y − x) + xy(y − x)
(
= (x − y) z(x + z) + z 2 − xy ]
= (x − y)(z(x − z) + y(z − x)]
= (x − y)(z − x)[ − z + y]
= (x − y)(y − z)(z − x)
Hence, Δ = (x − y)(y − z)(z − x).
Page : 126 , Block Name : Exercise 4.4
| |
a 11 a 12 a 13
Q5 If Δ = a 21 a 22 a 23 and A is Cofactors of a , then value of Δ
ij ij
a 31 a 32 a 33
(A) a 11A 31 + a 12A 32 + a 13A 33 (B) a 11A 11 + a 12A 21 + a 13A 31
(C) a 21A 11 + a 22A 12 + a 23A 13 (D) a 11A 11 + a 21A 21 + a 31A 31
Answer.
Answer is : D
We know that:
Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors
∴ Δ = a 11A 11 + a 21A 21 + a 31A 31
Hence, the value of Δ is given by the expression given in alternative D.
The correct answer is D.
1
x
z
=z−x
M 33 = | | 1 x
1 y
=y−x
∴ A 13 = cofactor of a 13 = ( − 1) 1 + 3M 13 = (z − y)
A 23 = cofactor of a 23 = ( − 1) 2 + 3M 23 = − (z − x) = (x − z)
A 33 = cofactor of a 33 = ( − 1) 3 + 3M 33 = (y − x)
We know that Δ is equal to the sum of the product of the elements of the second row with their
corresponding cofactors.
∴ Δ = a 13A 13 + a 23A 23 + a 33A 33
= yz(z − y) + zx(x − z) + xy(y − x)
= yz 2 − y 2z + x 2z − xz 2 + xy 2 − x 2y
( ) ( ) (
= x 2z − y 2z + yz 2 − xz 2 + xy 2 − x 2y )
= z (x − y ) + z (y − x) + xy(y − x)
2 2
= z(x − y)(x + y) + z 2(y − x) + xy(y − x)
(
= (x − y) z(x + z) + z 2 − xy ]
= (x − y)(z(x − z) + y(z − x)]
= (x − y)(z − x)[ − z + y]
= (x − y)(y − z)(z − x)
Hence, Δ = (x − y)(y − z)(z − x).
Page : 126 , Block Name : Exercise 4.4
| |
a 11 a 12 a 13
Q5 If Δ = a 21 a 22 a 23 and A is Cofactors of a , then value of Δ
ij ij
a 31 a 32 a 33
(A) a 11A 31 + a 12A 32 + a 13A 33 (B) a 11A 11 + a 12A 21 + a 13A 31
(C) a 21A 11 + a 22A 12 + a 23A 13 (D) a 11A 11 + a 21A 21 + a 31A 31
Answer.
Answer is : D
We know that:
Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors
∴ Δ = a 11A 11 + a 21A 21 + a 31A 31
Hence, the value of Δ is given by the expression given in alternative D.
The correct answer is D.
Page 33
Page : 126 , Block Name : Exercise 4.4
Q1 Find adjoint of the matrix.
[ ] 1 2
3 4
Answer.
Let A =
[ ]
1 2
3 4
We have,
A 11 = 4, A 12 = − 3, A 21 = − 2, A 22 = 1
[ ][ ]
A 11 A 21 4 −2
∴ adjA = =
A 12 A 22 −3 1
Page : 131 , Block Name : Exercise 4.5
Q2 Find adjoint of the matrix.
[ ]
1 −1 2
2 3 5
−2 0 1
[ ]
1 −1 2
Answer. A = 2 3 5
−2 0 1
A 11 =
| |3
0
5
1
=3−0=3
We have, A 12 = −
| | 2
−2
5
1
= − (2 + 10) = − 12
A 13 = | |2
−2
3
0
=0+6=6
Q1 Find adjoint of the matrix.
[ ] 1 2
3 4
Answer.
Let A =
[ ]
1 2
3 4
We have,
A 11 = 4, A 12 = − 3, A 21 = − 2, A 22 = 1
[ ][ ]
A 11 A 21 4 −2
∴ adjA = =
A 12 A 22 −3 1
Page : 131 , Block Name : Exercise 4.5
Q2 Find adjoint of the matrix.
[ ]
1 −1 2
2 3 5
−2 0 1
[ ]
1 −1 2
Answer. A = 2 3 5
−2 0 1
A 11 =
| |3
0
5
1
=3−0=3
We have, A 12 = −
| | 2
−2
5
1
= − (2 + 10) = − 12
A 13 = | |2
−2
3
0
=0+6=6
Page 34
A 21 = − | | −1
0
2
1
= − ( − 1 − 0) = 1
A 22 = | |
−2
1 2
1
=1+4=5
A 23 = −
| | 1
−2
−1
0
= − (0 − 2) = 2
A 31 =
| |
−1
3
2
5
= − 5 − 6 = − 11
A 32 = −
| | 1 2
2 5
= − (5 − 4) = − 1
A 33 =
| |
1
2
−1
3
=3+2=5
[ ][ ]
A 11 A 21 A 31
3 1 − 11
Hence, adjA = A 12 A 22 A 32 = − 12 5 −1
A 13 A 23 A 33 6 2 5
Page : 131 , Block Name : Exercise 4.5
Q3 A(adjA) = (adjA)A = | A | I
[ 2
−4
3
−6 ]
Answer. A= [ −4
2
−6
3
]
we have,
| A | = − 12 − ( − 12) = − 12 + 12 = 0
∴ |A|I = 0 [ ] [ ]
1
0
0
1
=
0
0
0
0
now, A 11 = − 6, A 12 = 4, A 21 = − 3, A 22 = 2
∴ adjA =
[ −6
4
−3
2 ]
0
2
1
= − ( − 1 − 0) = 1
A 22 = | |
−2
1 2
1
=1+4=5
A 23 = −
| | 1
−2
−1
0
= − (0 − 2) = 2
A 31 =
| |
−1
3
2
5
= − 5 − 6 = − 11
A 32 = −
| | 1 2
2 5
= − (5 − 4) = − 1
A 33 =
| |
1
2
−1
3
=3+2=5
[ ][ ]
A 11 A 21 A 31
3 1 − 11
Hence, adjA = A 12 A 22 A 32 = − 12 5 −1
A 13 A 23 A 33 6 2 5
Page : 131 , Block Name : Exercise 4.5
Q3 A(adjA) = (adjA)A = | A | I
[ 2
−4
3
−6 ]
Answer. A= [ −4
2
−6
3
]
we have,
| A | = − 12 − ( − 12) = − 12 + 12 = 0
∴ |A|I = 0 [ ] [ ]
1
0
0
1
=
0
0
0
0
now, A 11 = − 6, A 12 = 4, A 21 = − 3, A 22 = 2
∴ adjA =
[ −6
4
−3
2 ]
Page 35
Now,
A(adjA) =
[ 2
−4
3
−6 ][ ] −6
4
−3
2
= [ − 12 + 12
24 − 24
−6 + 6
12 − 12 ] [ ] =
0
0
0
0
(adjA)A = [ −6
4
−3
2 ][ ] 2
−4
3
−6
Also,
= [ − 12 + 12
8−8
− 18 + 18
12 − 12 ] [ ]
=
0 0
0 0
Hence, A(adjA) = (adjA)A = | A | I
Page : 131 , Block Name : Exercise 4.5
Q4 A(adjA) = (adjA)A = | A | I
[ ]
1 −1 2
3 0 −2
1 0 3
[ ]
1 −1 2
Answer. A= 3 0 −2
1 0 3
| A | = 1(0 − 0) + 1(9 + 2) + 2(0 − 0) = 11
[ ][ ]
1 0 0 11 0 0
∴ | A | I = 11 0 1 0 = 0 11 0
0 0 1 0 0 11
Now,
A 11 = 0, A 12 = − (9 + 2) = − 11, A 13 = 0
A 21 = − ( − 3 − 0) = 3, A 22 = 3 − 2 = 1, A 23 = − (0 + 1) = − 1
A 31 = 2 − 0 = 2, A 32 = − ( − 2 − 6) = 8, A 33 = 0 + 3 = 3
[ ]
0 3 2
∴ adjA = − 11 1 8
0 −1 3
[ ][ ]
1 −1 2 0 3 2
A(adjA) = 3 0 −2 − 11 1 8
1 0 3 0 −1 3
A(adjA) =
[ 2
−4
3
−6 ][ ] −6
4
−3
2
= [ − 12 + 12
24 − 24
−6 + 6
12 − 12 ] [ ] =
0
0
0
0
(adjA)A = [ −6
4
−3
2 ][ ] 2
−4
3
−6
Also,
= [ − 12 + 12
8−8
− 18 + 18
12 − 12 ] [ ]
=
0 0
0 0
Hence, A(adjA) = (adjA)A = | A | I
Page : 131 , Block Name : Exercise 4.5
Q4 A(adjA) = (adjA)A = | A | I
[ ]
1 −1 2
3 0 −2
1 0 3
[ ]
1 −1 2
Answer. A= 3 0 −2
1 0 3
| A | = 1(0 − 0) + 1(9 + 2) + 2(0 − 0) = 11
[ ][ ]
1 0 0 11 0 0
∴ | A | I = 11 0 1 0 = 0 11 0
0 0 1 0 0 11
Now,
A 11 = 0, A 12 = − (9 + 2) = − 11, A 13 = 0
A 21 = − ( − 3 − 0) = 3, A 22 = 3 − 2 = 1, A 23 = − (0 + 1) = − 1
A 31 = 2 − 0 = 2, A 32 = − ( − 2 − 6) = 8, A 33 = 0 + 3 = 3
[ ]
0 3 2
∴ adjA = − 11 1 8
0 −1 3
[ ][ ]
1 −1 2 0 3 2
A(adjA) = 3 0 −2 − 11 1 8
1 0 3 0 −1 3
Page 36
[ ]
0+9+2 0+0+0 0−6+6
= − 11 + 3 + 8 11 + 0 + 0 − 22 − 2 + 24
0−3+3 0+0+0 0+2+9
[ ]
11 0 0
= 0 11 0
0 0 11
Hence,A(adjA) = (adjA)A = | A | I
Page : 131 , Block Name : Exercise 4.5
Q5 Find the inverse of the matrix
[ ]
2
4
−2
3
Answer. Let, A=
[ ]
2
4
−2
3
We have,
| A | = − 2 + 15 = 13
Now,
A 11 = 2, A 12 = 3, A 21 = − 5, A 22 = − 1
∴ adjA =
[ ] 2
3
−5
−1
1
∴ A − 1 = | A | adjA = 13
1
[ ] 2
3
−5
−1
Page : 132 , Block Name : Exercise 4.5
Q6 Find the inverse of the matrix
[ ]
−1
−3
5
2
Answer. Let A= [ ]
−1
−3
5
2
We have,
| A | = − 2 + 15 = 13
Now
A 11 = 2, A 12 = 3, A 21 = − 5, A 22 = − 1
0+9+2 0+0+0 0−6+6
= − 11 + 3 + 8 11 + 0 + 0 − 22 − 2 + 24
0−3+3 0+0+0 0+2+9
[ ]
11 0 0
= 0 11 0
0 0 11
Hence,A(adjA) = (adjA)A = | A | I
Page : 131 , Block Name : Exercise 4.5
Q5 Find the inverse of the matrix
[ ]
2
4
−2
3
Answer. Let, A=
[ ]
2
4
−2
3
We have,
| A | = − 2 + 15 = 13
Now,
A 11 = 2, A 12 = 3, A 21 = − 5, A 22 = − 1
∴ adjA =
[ ] 2
3
−5
−1
1
∴ A − 1 = | A | adjA = 13
1
[ ] 2
3
−5
−1
Page : 132 , Block Name : Exercise 4.5
Q6 Find the inverse of the matrix
[ ]
−1
−3
5
2
Answer. Let A= [ ]
−1
−3
5
2
We have,
| A | = − 2 + 15 = 13
Now
A 11 = 2, A 12 = 3, A 21 = − 5, A 22 = − 1
Page 37
∴ adjA = [ ] 2
3
−5
−1
1
∴ A − 1 = | A | adjA = 13
1
[ ] 2
3
−5
−1
Page : 132 , Block Name : Exercise 4.5
Q7 Find the inverse of the matrix
[ ]
1 2 3
0 2 4
0 0 5
[ ]
1 2 3
Answer. Let A= 0 2 4 we have,
0 0 5
| A | = 1(10 − 0) − 2(0 − 0) + 3(0 − 0) = 10
Now,
A 11 = 10 − 0 = 10, A 12 = − (0 − 0) = 0, A 13 = 0 − 0 = 0
A 21 = − (10 − 0) = − 10, A 22 = 5 − 0 = 5, A 23 = − (0 − 0) = 0
A 31 = 8 − 6 = 2, A 12 = − (4 − 0) = − 4, A 33 = 2 − 0 = 2
[ ]
10 − 10 2
∴ adjA = 0 5 −4
0 0 2
[ ]
10 − 10 2
1 1
∴ A − 1 = | A | adjA = 10 0 5 −4
0 0 2
Page : 132 , Block Name : Exercise 4.5
Q8 Find the inverse of the matrix
[ ]
1 0 0
3 3 0
5 2 −1
3
−5
−1
1
∴ A − 1 = | A | adjA = 13
1
[ ] 2
3
−5
−1
Page : 132 , Block Name : Exercise 4.5
Q7 Find the inverse of the matrix
[ ]
1 2 3
0 2 4
0 0 5
[ ]
1 2 3
Answer. Let A= 0 2 4 we have,
0 0 5
| A | = 1(10 − 0) − 2(0 − 0) + 3(0 − 0) = 10
Now,
A 11 = 10 − 0 = 10, A 12 = − (0 − 0) = 0, A 13 = 0 − 0 = 0
A 21 = − (10 − 0) = − 10, A 22 = 5 − 0 = 5, A 23 = − (0 − 0) = 0
A 31 = 8 − 6 = 2, A 12 = − (4 − 0) = − 4, A 33 = 2 − 0 = 2
[ ]
10 − 10 2
∴ adjA = 0 5 −4
0 0 2
[ ]
10 − 10 2
1 1
∴ A − 1 = | A | adjA = 10 0 5 −4
0 0 2
Page : 132 , Block Name : Exercise 4.5
Q8 Find the inverse of the matrix
[ ]
1 0 0
3 3 0
5 2 −1
Page 38
[ ]
1 0 0
Answer. Let A= 3 3 0
5 2 −1
We have,
| A | = 1( − 3 − 0) − 0 + 0 = − 3
Now,
A 11 = − 3 − 0 = − 3, A 12 = − ( − 3 − 0) = 3, A 13 = 6 − 15 = − 9
A 21 = − (0 − 0) = 0, A 22 = − 1 − 0 = − 1, A 23 = − (2 − 0) = − 2
A 31 = 0 − 0 = 0, A 32 = − (0 − 0) = 0, A 33 = 3 − 0 = 3
[ ]
−3 0 0
∴ adjA = 3 −1 0
−9 −2 3
[ ]
−3 0 0
1 1
∴ A − 1 = | A | adj A = − 3 3 −1 0
−9 −2 3
Page : 132 , Block Name : Exercise 4.5
Q9 Find the inverse of the matrix
[ ]
2 1 3
4 −1 0
−7 2 1
[ ]
2 1 3
Answer. Let A= 4 −1 0
−7 2 1
We have,
| A | = 2( − 1 − 0) − 1(4 − 0) + 3(8 − 7)
= 2( − 1) − 1(4) + 3(1)
= −2−4+3
= −3
now, A 11 = − 1 − 0 = − 1, A 12 = − (4 − 0) = − 4, A 13 = 8 − 7 = 1
A 21 = − (1 − 6) = 5, A 22 = 2 + 21 = 23, A 13 = − (4 + 7) = − 11
A 31 = 0 + 3 = 3, A 12 = − (0 − 12) = 12, A 33 = − 2 − 4 = − 6
1 0 0
Answer. Let A= 3 3 0
5 2 −1
We have,
| A | = 1( − 3 − 0) − 0 + 0 = − 3
Now,
A 11 = − 3 − 0 = − 3, A 12 = − ( − 3 − 0) = 3, A 13 = 6 − 15 = − 9
A 21 = − (0 − 0) = 0, A 22 = − 1 − 0 = − 1, A 23 = − (2 − 0) = − 2
A 31 = 0 − 0 = 0, A 32 = − (0 − 0) = 0, A 33 = 3 − 0 = 3
[ ]
−3 0 0
∴ adjA = 3 −1 0
−9 −2 3
[ ]
−3 0 0
1 1
∴ A − 1 = | A | adj A = − 3 3 −1 0
−9 −2 3
Page : 132 , Block Name : Exercise 4.5
Q9 Find the inverse of the matrix
[ ]
2 1 3
4 −1 0
−7 2 1
[ ]
2 1 3
Answer. Let A= 4 −1 0
−7 2 1
We have,
| A | = 2( − 1 − 0) − 1(4 − 0) + 3(8 − 7)
= 2( − 1) − 1(4) + 3(1)
= −2−4+3
= −3
now, A 11 = − 1 − 0 = − 1, A 12 = − (4 − 0) = − 4, A 13 = 8 − 7 = 1
A 21 = − (1 − 6) = 5, A 22 = 2 + 21 = 23, A 13 = − (4 + 7) = − 11
A 31 = 0 + 3 = 3, A 12 = − (0 − 12) = 12, A 33 = − 2 − 4 = − 6
Page 39
[ ]
−1 5 3
∴ adjA = −4 23 12
1 − 11 −6
[ ]
−1 5 3
1 1
∴ A − 1 = | A | adj A = − 3 − 4 23 12
1 − 11 −6
Page : 132 , Block Name : Exercise 4.5
[ ]
1 −1 2
Q10 Find the inverse of the matrix 0 2 −3
3 −2 4
[ ]
1 −1 2
Answer. Let A= 0 2 −3
3 −2 4
By expanding along C1, we have :
| A | = 1(8 − 6) − 0 + 3(3 − 4) = 2 − 3 = − 1
Now,
A 11 = 8 − 6 = 2, A 12 = − (0 + 9) = − 9, A 3 = 0 − 6 = − 6
A 21 = − ( − 4 + 4) = 0, A 22 = 4 − 6 = − 2, A 23 = − ( − 2 + 3) = − 1
A 31 = 3 − 4 = − 1, A 12 = − ( − 3 − 0) = 3, A 33 = 2 − 0 = 2
[ ]
2 0 −1
∴ adjA = −9 −2 3
−6 −1 2
[ ][ ]
2 0 −1 −2 0 −3
1
∴ A − 1 = | A | adjA = − −9 −2 3 = 9 2 −3
−6 −1 2 6 1 −2
Page : 132 , Block Name : Exercise 4.5
Q11 Find the inverse of the matrix
[ ]
1 0 0
0 cosα sinα
0 sinα − cosα
−1 5 3
∴ adjA = −4 23 12
1 − 11 −6
[ ]
−1 5 3
1 1
∴ A − 1 = | A | adj A = − 3 − 4 23 12
1 − 11 −6
Page : 132 , Block Name : Exercise 4.5
[ ]
1 −1 2
Q10 Find the inverse of the matrix 0 2 −3
3 −2 4
[ ]
1 −1 2
Answer. Let A= 0 2 −3
3 −2 4
By expanding along C1, we have :
| A | = 1(8 − 6) − 0 + 3(3 − 4) = 2 − 3 = − 1
Now,
A 11 = 8 − 6 = 2, A 12 = − (0 + 9) = − 9, A 3 = 0 − 6 = − 6
A 21 = − ( − 4 + 4) = 0, A 22 = 4 − 6 = − 2, A 23 = − ( − 2 + 3) = − 1
A 31 = 3 − 4 = − 1, A 12 = − ( − 3 − 0) = 3, A 33 = 2 − 0 = 2
[ ]
2 0 −1
∴ adjA = −9 −2 3
−6 −1 2
[ ][ ]
2 0 −1 −2 0 −3
1
∴ A − 1 = | A | adjA = − −9 −2 3 = 9 2 −3
−6 −1 2 6 1 −2
Page : 132 , Block Name : Exercise 4.5
Q11 Find the inverse of the matrix
[ ]
1 0 0
0 cosα sinα
0 sinα − cosα
Page 40
[ ]
1 0 0
Answer. Let A= 0 cosα sinα
0 sinα − cosα
We have,
( )
| A | = 1 − cos 2α − sin 2α = − cos 2α + sin 2α = − 1 ( )
Now,
A 11 = − cos 2α − sin 2α = − 1, A 12 = 0, A 13 = 0
A 21 = 0, A 22 = − cosα, A 23 = − sinα
A 31 = 0, A 32 = − sinα, A 33 = cosα
[ ]
−1 0 0
∴ adjA = 0 − cosα − sinα
0 − sinα cosα
[ ][ ]
−1 0 0 1 0 0
1
∴ A − 1 = | A | ⋅ adjA = − 0 − cosα − sinα = 0 cosα sinα
0 − sinα cosα 0 sinα − cosα
Page : 132 , Block Name : Exercise 4.5
Q12 Let A = [ ] [ ]
3
2
7
5 and
6 =
6
7
8
9
. Verify that (AB) − 1 = B − 1A − 1
Answer. Let A=
[ ] 3
2
7
5
We have,
| A | = 15 − 14 = 1
Now,
A 11 = 5, A 12 = − 2, A 21 = − 7, A 22 = 3
∴ adjA = [ 5
−2
−7
3 ]
1
∴ A − 1 = | A | ⋅ adjA = [ 5
−2
−7
3 ]
1 0 0
Answer. Let A= 0 cosα sinα
0 sinα − cosα
We have,
( )
| A | = 1 − cos 2α − sin 2α = − cos 2α + sin 2α = − 1 ( )
Now,
A 11 = − cos 2α − sin 2α = − 1, A 12 = 0, A 13 = 0
A 21 = 0, A 22 = − cosα, A 23 = − sinα
A 31 = 0, A 32 = − sinα, A 33 = cosα
[ ]
−1 0 0
∴ adjA = 0 − cosα − sinα
0 − sinα cosα
[ ][ ]
−1 0 0 1 0 0
1
∴ A − 1 = | A | ⋅ adjA = − 0 − cosα − sinα = 0 cosα sinα
0 − sinα cosα 0 sinα − cosα
Page : 132 , Block Name : Exercise 4.5
Q12 Let A = [ ] [ ]
3
2
7
5 and
6 =
6
7
8
9
. Verify that (AB) − 1 = B − 1A − 1
Answer. Let A=
[ ] 3
2
7
5
We have,
| A | = 15 − 14 = 1
Now,
A 11 = 5, A 12 = − 2, A 21 = − 7, A 22 = 3
∴ adjA = [ 5
−2
−7
3 ]
1
∴ A − 1 = | A | ⋅ adjA = [ 5
−2
−7
3 ]
Page 41
Now, let B =
[ ] 6
7
8
9
.
We have,
| B | = 54 − 56 = − 2
[ ]
9
−2
[ ]
9 −8 4
∴ adj B = = 7
−7 6
2 −3
[ ][ ]
9
−2 4 5 −7
B − 1A − 1 = 7 −2 3
2 −3
Now, …(i)Then,
[ ][ ]
45 63 61 87
− 2 −8 2
+ 12 − 2 2
= 35 49
= 47 67
2
+6 − 2 −9 2
− 2
AB =
[ ][ ]
3
2
7
5
6
2
8
9
= [ ]
18 + 49
12 + 35
24 + 63
16 + 45
= [ ] 67
47
87
61
Therefore, we have | AB | = 67 × 61 − 87 × 47 = 4087 − 4089 = − 2.
Also,
adj(AB) = [ 61
− 47
− 87
67 ]
∴ (AB) − 1 =
1
| AB |
adj(AB) = −
1
[
2 − 47
61 − 87
67 ]
[ ]
61 87
− 2 2
= 47 67
…(ii)
2
− 2
[ ] 6
7
8
9
.
We have,
| B | = 54 − 56 = − 2
[ ]
9
−2
[ ]
9 −8 4
∴ adj B = = 7
−7 6
2 −3
[ ][ ]
9
−2 4 5 −7
B − 1A − 1 = 7 −2 3
2 −3
Now, …(i)Then,
[ ][ ]
45 63 61 87
− 2 −8 2
+ 12 − 2 2
= 35 49
= 47 67
2
+6 − 2 −9 2
− 2
AB =
[ ][ ]
3
2
7
5
6
2
8
9
= [ ]
18 + 49
12 + 35
24 + 63
16 + 45
= [ ] 67
47
87
61
Therefore, we have | AB | = 67 × 61 − 87 × 47 = 4087 − 4089 = − 2.
Also,
adj(AB) = [ 61
− 47
− 87
67 ]
∴ (AB) − 1 =
1
| AB |
adj(AB) = −
1
[
2 − 47
61 − 87
67 ]
[ ]
61 87
− 2 2
= 47 67
…(ii)
2
− 2
Page 42
From (1) and (2), we have:
(AB) − 1 = B − 1A − 1
Hence, the given result is proved.
Page : 132 , Block Name : Exercise 4.5
Q13 If A = [ ] 3
−1
1
2
, show that A 2 − 5A + 7I = O . Hence find A − 1
Answer.
A=
[ ]
−1
3 1
2
[ ][ ] [
A2 = A ⋅ A =
3
−1
1
2
3
−1
1
2
=
9−1
−3 − 2
3+2
−1 + 4 ] [ ]
=
8
−5
5
3
∴ A 2 − 5A + 7I
=
[ ] [ ] [ ]
8
−5
5
3
−5
3
−1
1
2
+7
1 0
0 1
= [ ][ ][ ]
8
−5
5
3
−
15
−5
5
5
+
7
0
0
7
= [ ] [ ][ ]
−7
0
0
−7
+
7
0
0
7
0
0
0
0
Hence, A 2 − 5A + 7I = O
∴ A ⋅ A − 5A = − 7I
( )
⇒ A ⋅ A A − 1 − 5AA − 1 = − 7IA − 1 [ Post-multiplying by A as | A | ≠ 0 ]
−1
⇒ A (AA ) − 5I = − 7A
−1 −1
⇒ AI − 5I = − 7A − 1
1
⇒ A − 1 = − 7 (A − 5I)
1
⇒ A − 1 = 7 (5I − A)
= 7
1
([ ] [ ])
5
0
0
5
−
−1
3 1
2
= 7
1
[ ]
2
1
−1
3
(AB) − 1 = B − 1A − 1
Hence, the given result is proved.
Page : 132 , Block Name : Exercise 4.5
Q13 If A = [ ] 3
−1
1
2
, show that A 2 − 5A + 7I = O . Hence find A − 1
Answer.
A=
[ ]
−1
3 1
2
[ ][ ] [
A2 = A ⋅ A =
3
−1
1
2
3
−1
1
2
=
9−1
−3 − 2
3+2
−1 + 4 ] [ ]
=
8
−5
5
3
∴ A 2 − 5A + 7I
=
[ ] [ ] [ ]
8
−5
5
3
−5
3
−1
1
2
+7
1 0
0 1
= [ ][ ][ ]
8
−5
5
3
−
15
−5
5
5
+
7
0
0
7
= [ ] [ ][ ]
−7
0
0
−7
+
7
0
0
7
0
0
0
0
Hence, A 2 − 5A + 7I = O
∴ A ⋅ A − 5A = − 7I
( )
⇒ A ⋅ A A − 1 − 5AA − 1 = − 7IA − 1 [ Post-multiplying by A as | A | ≠ 0 ]
−1
⇒ A (AA ) − 5I = − 7A
−1 −1
⇒ AI − 5I = − 7A − 1
1
⇒ A − 1 = − 7 (A − 5I)
1
⇒ A − 1 = 7 (5I − A)
= 7
1
([ ] [ ])
5
0
0
5
−
−1
3 1
2
= 7
1
[ ]
2
1
−1
3
Page 43
∴ A −1 = 7
1
[ ] 2
1
−1
3
Page : 132 , Block Name : Exercise 4.5
Q14 For the matrix A = [ ]
3
1
2
1
, find the numbers a and b such that A 2 + aA + bI = 0.
Answer.
A=
[ ]
3
1
2
1
∴ A2 =
[ ][ ] [
3
1
2
1
3
1
2
1
=
9+2
3+1
6+2
2+1 ] [ ]
=
11
4
8
3
Now,
A 2 + aA + bl = O
⇒ (AA)A − 1 + aAA − 1 + bIA − 1 = O
[ Post-multiplying by A as | A | ≠ 0 ]
−1
( )
⇒ A AA − 1 + aI + b LA − 1 = O ( )
⇒ AI + al + bA − 1 = O
⇒ A + al = − bA − 1
1
⇒ A − 1 = − b (A + aI)
Now,
1
A − 1 = | A | adjA = 1
1
[ 1
−1
−2
3 ] [ =
1
−1
−2
3 ]
We have:
[ ]
−3−a 2
−b
[ 1
−1
−2
3 ] ([ ] [ ])
= − b
1 3
1
2
1
+
a 0
0 a
= − b
1
[ 3+a
1
2
1+a ] =
b
−b
1 −1−a
b
Comparing the corresponding elements of the two matrices, we have:
1
−b = −1 ⇒ b = 1
−3−a
b
=1⇒ −3−a=1⇒a= −4
Hence, − 4 and 1 are the required values of a and b respectively.
Page : 132 , Block Name : Exercise 4.5
1
[ ] 2
1
−1
3
Page : 132 , Block Name : Exercise 4.5
Q14 For the matrix A = [ ]
3
1
2
1
, find the numbers a and b such that A 2 + aA + bI = 0.
Answer.
A=
[ ]
3
1
2
1
∴ A2 =
[ ][ ] [
3
1
2
1
3
1
2
1
=
9+2
3+1
6+2
2+1 ] [ ]
=
11
4
8
3
Now,
A 2 + aA + bl = O
⇒ (AA)A − 1 + aAA − 1 + bIA − 1 = O
[ Post-multiplying by A as | A | ≠ 0 ]
−1
( )
⇒ A AA − 1 + aI + b LA − 1 = O ( )
⇒ AI + al + bA − 1 = O
⇒ A + al = − bA − 1
1
⇒ A − 1 = − b (A + aI)
Now,
1
A − 1 = | A | adjA = 1
1
[ 1
−1
−2
3 ] [ =
1
−1
−2
3 ]
We have:
[ ]
−3−a 2
−b
[ 1
−1
−2
3 ] ([ ] [ ])
= − b
1 3
1
2
1
+
a 0
0 a
= − b
1
[ 3+a
1
2
1+a ] =
b
−b
1 −1−a
b
Comparing the corresponding elements of the two matrices, we have:
1
−b = −1 ⇒ b = 1
−3−a
b
=1⇒ −3−a=1⇒a= −4
Hence, − 4 and 1 are the required values of a and b respectively.
Page : 132 , Block Name : Exercise 4.5
Page 44
[ ]
1 1 1
Q15 For the matrix A = 1 2 −3 = 0.
2 −1 3 show that A 3 − 6A 2 + 5A + 11
hence ,A − 1
Answer.
[ ]
1 1 1
A= 1 2 −3
2 −1 3
[ ][ ]
1 1 1 1 1 1
A2 = 1 2 −3 1 2 −3
2 −1 3 2 −1 3
[ ][ ]
1+1+2 1+2−1 1−3+3 4 2 1
= 1+2−6 1+4+3 1−6−9 = −3 8 − 14
2−1+6 2−2−3 2+3+9 7 −3 14
[ ][ ]
4 2 1 1 1 1
A3 = A2 ⋅ A = −3 8 − 14 1 2 −3
7 −3 14 2 −1 3
[ ]
4+2+2 4+4−1 4−6+3
= − 3 + 8 − 28 − 3 + 16 + 14 − 3 − 24 − 42
7 − 3 + 28 7 − 6 − 14 7 + 9 + 42
[ ]
8 7 − 69
= − 23 27 − 69
32 − 13 58
∴ A 3 − 6A 2 + 5A + 11I
[ ][ ][ ] [ ]
8 7 1 4 2 1 1 1 1 1 0 0
= − 23 27 − 69 − 6 −3 8 − 14 +5 1 2 −3 + 11 0 1 0
32 − 13 58 7 −3 14 2 −1 3 0 0 1
[ ][ ][ ]
4 2 1 6 6 6 5 0 0
= −3 8 − 14 − 6 12 − 18 + 0 5 0
7 −3 14 12 −6 18 0 0 5
1 1 1
Q15 For the matrix A = 1 2 −3 = 0.
2 −1 3 show that A 3 − 6A 2 + 5A + 11
hence ,A − 1
Answer.
[ ]
1 1 1
A= 1 2 −3
2 −1 3
[ ][ ]
1 1 1 1 1 1
A2 = 1 2 −3 1 2 −3
2 −1 3 2 −1 3
[ ][ ]
1+1+2 1+2−1 1−3+3 4 2 1
= 1+2−6 1+4+3 1−6−9 = −3 8 − 14
2−1+6 2−2−3 2+3+9 7 −3 14
[ ][ ]
4 2 1 1 1 1
A3 = A2 ⋅ A = −3 8 − 14 1 2 −3
7 −3 14 2 −1 3
[ ]
4+2+2 4+4−1 4−6+3
= − 3 + 8 − 28 − 3 + 16 + 14 − 3 − 24 − 42
7 − 3 + 28 7 − 6 − 14 7 + 9 + 42
[ ]
8 7 − 69
= − 23 27 − 69
32 − 13 58
∴ A 3 − 6A 2 + 5A + 11I
[ ][ ][ ] [ ]
8 7 1 4 2 1 1 1 1 1 0 0
= − 23 27 − 69 − 6 −3 8 − 14 +5 1 2 −3 + 11 0 1 0
32 − 13 58 7 −3 14 2 −1 3 0 0 1
[ ][ ][ ]
4 2 1 6 6 6 5 0 0
= −3 8 − 14 − 6 12 − 18 + 0 5 0
7 −3 14 12 −6 18 0 0 5
Page 45
[ ][ ]
9 2 1 6 6 6
= −3 13 − 14 − 6 12 − 18
7 −3 19 12 −6 18
[ ]
3 −4 −5
= −9 1 4
−5 3 1
From equation (1), we have:
[ ] [ ]
3 −4 −5 −3 4 5
1 1
A − 1 = − 11 − 9 1 4 = 11 9 −1 −4
−5 3 1 5 −3 −1
Page : 132 , Block Name : Exercise 4.5
[ ]
2 −1 1
Q16 If A = −1 2 −1
1 −1 2 verify that A 3 − 6A 2 + 9A − 4I = O and hence find A − 1
Answer.
[ ]
2 −1 1
A= −1 2 −1
1 −1 2
[ ][ ]
2 −1 1 2 −1 1
A2 = −1 2 −1 −1 2 −1
1 −1 2 1 −1 2
[ ]
4+1+1 −2 − 2 − 1 2+1+2
= −2 − 2 − 1 1+4+1 −1 − 2 − 2
2+1+2 −1 − 2 − 2 1+1+4
[ ]
6 −5 5
= −5 6 −5
5 −5 6
[ ][ ]
6 −5 5 2 −1 1
A 3 = A 2A = −5 6 −5 −1 2 −1
5 −5 6 1 −1 2
9 2 1 6 6 6
= −3 13 − 14 − 6 12 − 18
7 −3 19 12 −6 18
[ ]
3 −4 −5
= −9 1 4
−5 3 1
From equation (1), we have:
[ ] [ ]
3 −4 −5 −3 4 5
1 1
A − 1 = − 11 − 9 1 4 = 11 9 −1 −4
−5 3 1 5 −3 −1
Page : 132 , Block Name : Exercise 4.5
[ ]
2 −1 1
Q16 If A = −1 2 −1
1 −1 2 verify that A 3 − 6A 2 + 9A − 4I = O and hence find A − 1
Answer.
[ ]
2 −1 1
A= −1 2 −1
1 −1 2
[ ][ ]
2 −1 1 2 −1 1
A2 = −1 2 −1 −1 2 −1
1 −1 2 1 −1 2
[ ]
4+1+1 −2 − 2 − 1 2+1+2
= −2 − 2 − 1 1+4+1 −1 − 2 − 2
2+1+2 −1 − 2 − 2 1+1+4
[ ]
6 −5 5
= −5 6 −5
5 −5 6
[ ][ ]
6 −5 5 2 −1 1
A 3 = A 2A = −5 6 −5 −1 2 −1
5 −5 6 1 −1 2
Page 46
[ ]
12 + 5 + 5 − 6 − 10 − 5 6 + 5 + 10
= − 10 − 6 − 5 5 + 12 + 5 − 5 − 6 − 10
10 + 5 + 6 − 5 − 10 − 6 5 + 5 + 12
[ ]
22 − 21 21
= − 21 22 − 21
21 − 21 22
now ,
A 3 − 6A 2 + 9A − 4I
[ ][ ][ ][ ]
22 − 21 21 6 −5 5 2 −1 1 1 0 0
= − 21 22 − 21 − 6 −5 6 −5 + 9 −1 2 −1 −4 0 1 0
21 − 21 22 5 −5 6 1 −1 2 0 0 1
[ ][ ][ ][ ]
22 − 21 21 36 − 30 30 18 −9 9 4 0 0
= − 21 22 − 21 − − 30 36 − 30 + −9 18 −9 − 0 4 0
21 − 21 22 30 − 30 36 9 −9 18 0 0 4
[ ][ ][ ]
40 − 30 30 40 − 30 30 0 0 0
= − 30 40 − 30 − − 30 40 − 30 = 0 0 0
30 − 30 40 30 − 30 40 0 0 0
∴ A 3 − 6A 2 + 9A − 4I = O
Now,
A 3 − 6A 2 + 9A − 4I = O
A 3 − 6A 2 + 9A − 4I = O
⇒ (AAA)A − 6(AA)A + 9AA − 4LA = O [
−1 −1 Post-multiplying by A as | A | ≠ 0 ]
−1 −1
−1
⇒ AA (AA ) − 6A (AA ) + 9 (AA ) = 4 (IA )
−1 −1 −1 −1
⇒ AAI − 6AI + 9I = 4A − 1
⇒ A 2 − 6A + 9I = 4A − 1
( )
1
⇒ A − 1 = 4 A 2 − 6A + 9I . . . (i)
A 2 − 6A + 9I
[ ][ ][ ]
6 −5 5 2 −1 1 0 0 0
= −5 6 −5 − 6 −1 2 −1 +9 0 0 0
5 −5 6 1 −1 2 0 0 0
[ ][ ][ ]
6 −5 5 12 −6 6 9 0 0
= −5 6 −5 − −6 12 −6 + 0 9 0
5 −5 6 6 −6 12 0 0 9
12 + 5 + 5 − 6 − 10 − 5 6 + 5 + 10
= − 10 − 6 − 5 5 + 12 + 5 − 5 − 6 − 10
10 + 5 + 6 − 5 − 10 − 6 5 + 5 + 12
[ ]
22 − 21 21
= − 21 22 − 21
21 − 21 22
now ,
A 3 − 6A 2 + 9A − 4I
[ ][ ][ ][ ]
22 − 21 21 6 −5 5 2 −1 1 1 0 0
= − 21 22 − 21 − 6 −5 6 −5 + 9 −1 2 −1 −4 0 1 0
21 − 21 22 5 −5 6 1 −1 2 0 0 1
[ ][ ][ ][ ]
22 − 21 21 36 − 30 30 18 −9 9 4 0 0
= − 21 22 − 21 − − 30 36 − 30 + −9 18 −9 − 0 4 0
21 − 21 22 30 − 30 36 9 −9 18 0 0 4
[ ][ ][ ]
40 − 30 30 40 − 30 30 0 0 0
= − 30 40 − 30 − − 30 40 − 30 = 0 0 0
30 − 30 40 30 − 30 40 0 0 0
∴ A 3 − 6A 2 + 9A − 4I = O
Now,
A 3 − 6A 2 + 9A − 4I = O
A 3 − 6A 2 + 9A − 4I = O
⇒ (AAA)A − 6(AA)A + 9AA − 4LA = O [
−1 −1 Post-multiplying by A as | A | ≠ 0 ]
−1 −1
−1
⇒ AA (AA ) − 6A (AA ) + 9 (AA ) = 4 (IA )
−1 −1 −1 −1
⇒ AAI − 6AI + 9I = 4A − 1
⇒ A 2 − 6A + 9I = 4A − 1
( )
1
⇒ A − 1 = 4 A 2 − 6A + 9I . . . (i)
A 2 − 6A + 9I
[ ][ ][ ]
6 −5 5 2 −1 1 0 0 0
= −5 6 −5 − 6 −1 2 −1 +9 0 0 0
5 −5 6 1 −1 2 0 0 0
[ ][ ][ ]
6 −5 5 12 −6 6 9 0 0
= −5 6 −5 − −6 12 −6 + 0 9 0
5 −5 6 6 −6 12 0 0 9
Page 47
[ ]
3 1 −1
= 1 3 1
−1 1 3
From equation (1), we have:
[ ]
3 1 −1
1
A −1 = 4 1 3 1
−1 1 3
Page : 132 , Block Name : Exercise 4.5
Q17 A be a nonsingular square matrix of order 3 × 3. Then | adjA| is equal to
2
A. | A | B. | A | C. | A | 3 D. | A |
Answer. B
We know that,
[ ]
|A| 0 0
(adjA)A = | A | I = 0 |A| 0
0 0 |A|
| |
|A| 0 0
⇒ | (adjA)A | = 0 |A| 0 .
0 0 |A|
| |
1 0 0
⇒ | adjA‖A | = | A | 3 0 1 0 = | A | 3(I)
0 0 1
∴ | adjA | = | A | 2
Hence, the correct answer is B .
Page : 132 , Block Name : Exercise 4.5
Q18 If A is an invertible matrix of order 2, then det A − 1 ( )
1
A.det(A) B. det ( A ) C. 1 D. 0
1
Answer. Since A is an invertible matrix, A − 1 exists and A − 1 = | A | adjA
3 1 −1
= 1 3 1
−1 1 3
From equation (1), we have:
[ ]
3 1 −1
1
A −1 = 4 1 3 1
−1 1 3
Page : 132 , Block Name : Exercise 4.5
Q17 A be a nonsingular square matrix of order 3 × 3. Then | adjA| is equal to
2
A. | A | B. | A | C. | A | 3 D. | A |
Answer. B
We know that,
[ ]
|A| 0 0
(adjA)A = | A | I = 0 |A| 0
0 0 |A|
| |
|A| 0 0
⇒ | (adjA)A | = 0 |A| 0 .
0 0 |A|
| |
1 0 0
⇒ | adjA‖A | = | A | 3 0 1 0 = | A | 3(I)
0 0 1
∴ | adjA | = | A | 2
Hence, the correct answer is B .
Page : 132 , Block Name : Exercise 4.5
Q18 If A is an invertible matrix of order 2, then det A − 1 ( )
1
A.det(A) B. det ( A ) C. 1 D. 0
1
Answer. Since A is an invertible matrix, A − 1 exists and A − 1 = | A | adjA
Page 48
As matrix A is of order 2, let A =
[ ] a
c
b
d
Then, | A | = ad − bc and adjA =
[ ] d
−c
−b
a
Now,
[ ]
d −b
1 |A| |A|
A − 1 = | A | adjA = −c a
|A| |A|
| |
d −b
|A| |A|
| |
∴ A −1 = −c a
|A| |A|
=
1
|A| 2
1
| −c
d −b
a |1
=
1
|A| 2
(ad − bc)
= ⋅ |A| = |A|
|A| 2
( )
1
∴ det A − 1 = det ( A )
Hence, the correct answer is B .
Page : 132 , Block Name : Exercise 4.5
Q1 Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Answer.
The given system of equations is:
x + 2y = 2
2x + 3y = 3
The given system of equations can be written in the form of AX = B , where
A=
[ ] []
1
2
2
3
,X =
x
y
and B =
[]
2
3
[ ] a
c
b
d
Then, | A | = ad − bc and adjA =
[ ] d
−c
−b
a
Now,
[ ]
d −b
1 |A| |A|
A − 1 = | A | adjA = −c a
|A| |A|
| |
d −b
|A| |A|
| |
∴ A −1 = −c a
|A| |A|
=
1
|A| 2
1
| −c
d −b
a |1
=
1
|A| 2
(ad − bc)
= ⋅ |A| = |A|
|A| 2
( )
1
∴ det A − 1 = det ( A )
Hence, the correct answer is B .
Page : 132 , Block Name : Exercise 4.5
Q1 Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Answer.
The given system of equations is:
x + 2y = 2
2x + 3y = 3
The given system of equations can be written in the form of AX = B , where
A=
[ ] []
1
2
2
3
,X =
x
y
and B =
[]
2
3
Page 49
Now,
| A | = 1(3) − 2(2) = 3 − 4 = − 1 ≠ 0
∴ A is non-singular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q2 Examine the consistency of the system of equations.
2x − y = 5
x+y=4
Answer.
2x − y = 5
x+y=4
The given system of equations can be written in the form of AX = B, where
A= [ ] []
2
1
−1
1
,X =
x
z
and B = [] 5
4
Now,
| A | = 2(1) − ( − 1)(1) = 2 + 1 = 3 ≠ 0
∴ A is non-sinqular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q3 Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Answer. The given system of equations is:
x + 3y = 5
2x + 6y = 8
The given system of equations can be written in the form of AX = B, where,
A= [ ] []
1
2
3
6
,X =
x
y
and B = []
5
8
| A | = 1(6) − 3(2) = 6 − 6 = 0
∴ A is a singular matrix.
| A | = 1(3) − 2(2) = 3 − 4 = − 1 ≠ 0
∴ A is non-singular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q2 Examine the consistency of the system of equations.
2x − y = 5
x+y=4
Answer.
2x − y = 5
x+y=4
The given system of equations can be written in the form of AX = B, where
A= [ ] []
2
1
−1
1
,X =
x
z
and B = [] 5
4
Now,
| A | = 2(1) − ( − 1)(1) = 2 + 1 = 3 ≠ 0
∴ A is non-sinqular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q3 Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Answer. The given system of equations is:
x + 3y = 5
2x + 6y = 8
The given system of equations can be written in the form of AX = B, where,
A= [ ] []
1
2
3
6
,X =
x
y
and B = []
5
8
| A | = 1(6) − 3(2) = 6 − 6 = 0
∴ A is a singular matrix.
Page 50
Now,
(adjA) = [ 6
−2
−3
1 ]
(adjA)B = [ 6
−2
−3
1 ][ ] [
5
8
=
30 − 24
− 10 + 8 ] [ ]
=
6
−2
≠O
Thus, the solution of the given system of equations does not exist.
Hence, the system of equations is inconsistent.
Page : 136 , Block Name : Exercise 4.6
Q4 Examine the consistency of the system of equations.
x+y+z=1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Answer. The given system of equations is :
x+y+z=1
2x + 3y + 2z = 2
ax + ay + 2az = 4
This system of equations can be written in the form AX = B, where
[ ] [] []
1 1 1 x 1
A= 2 3 2 ,X = y and B = 2
a a 2a z 4
| A | = 1(6a − 2a) − 1(4a − 2a) + 1(2a − 3a)
= 4a − 2a − a = 4a − 3a = a ≠ 0
∴ A is non-singular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q5 Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = − 1
3x − 5y = 3
Answer. The given system of equations is:
3x − y − 2z = 2
2y − z = − 1
3x − 5y = 3
This system of equations can be written in the form of AX = B, where
(adjA) = [ 6
−2
−3
1 ]
(adjA)B = [ 6
−2
−3
1 ][ ] [
5
8
=
30 − 24
− 10 + 8 ] [ ]
=
6
−2
≠O
Thus, the solution of the given system of equations does not exist.
Hence, the system of equations is inconsistent.
Page : 136 , Block Name : Exercise 4.6
Q4 Examine the consistency of the system of equations.
x+y+z=1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Answer. The given system of equations is :
x+y+z=1
2x + 3y + 2z = 2
ax + ay + 2az = 4
This system of equations can be written in the form AX = B, where
[ ] [] []
1 1 1 x 1
A= 2 3 2 ,X = y and B = 2
a a 2a z 4
| A | = 1(6a − 2a) − 1(4a − 2a) + 1(2a − 3a)
= 4a − 2a − a = 4a − 3a = a ≠ 0
∴ A is non-singular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q5 Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = − 1
3x − 5y = 3
Answer. The given system of equations is:
3x − y − 2z = 2
2y − z = − 1
3x − 5y = 3
This system of equations can be written in the form of AX = B, where
Page 51
Now,
| A | = 3(0 − 5) − 0 + 3(1 + 4) = − 15 + 15 = 0
∴ A is a singular matrix.
Now,
(adjA) =
[ −5
−6 12
6 3
6 ]
[ ][ ] [ ][]
−5 10 5 2 − 10 − 10 + 15 −5
∴ (adjA)B = −3 6 3 −1 = −6 − 6 + 9 = −3 ≠O
−6 12 6 3 − 12 − 12 + 18 −6
Thus, the solution of the given system of equations does not exist.
Hence, the system of equations is inconsistent.
Page : 136 , Block Name : Exercise 4.6
Q6 Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = − 1
Answer. The given system of equations is:
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = − 1
[ ] [] [] [ ]
5 −1 4 x 5 5
A= 2 3 5 ,X = y and B = 2 and B = 2
5 −2 6 z z −1
| A | = 5(18 + 10) + 1(12 − 25) + 4( − 4 − 15)
= 5(28) + 1( − 13) + 4( − 19)
= 140 − 13 − 76
∴ A is non-singular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q7 Solve system of linear equations, using matrix method.
5x + 2y = 4
7x + 3y = 5
Answer. The given system of equations can be written in the form of AX = B, where
| A | = 3(0 − 5) − 0 + 3(1 + 4) = − 15 + 15 = 0
∴ A is a singular matrix.
Now,
(adjA) =
[ −5
−6 12
6 3
6 ]
[ ][ ] [ ][]
−5 10 5 2 − 10 − 10 + 15 −5
∴ (adjA)B = −3 6 3 −1 = −6 − 6 + 9 = −3 ≠O
−6 12 6 3 − 12 − 12 + 18 −6
Thus, the solution of the given system of equations does not exist.
Hence, the system of equations is inconsistent.
Page : 136 , Block Name : Exercise 4.6
Q6 Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = − 1
Answer. The given system of equations is:
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = − 1
[ ] [] [] [ ]
5 −1 4 x 5 5
A= 2 3 5 ,X = y and B = 2 and B = 2
5 −2 6 z z −1
| A | = 5(18 + 10) + 1(12 − 25) + 4( − 4 − 15)
= 5(28) + 1( − 13) + 4( − 19)
= 140 − 13 − 76
∴ A is non-singular.
Therefore, A − 1 exists.
Hence, the given system of equations is consistent.
Page : 136 , Block Name : Exercise 4.6
Q7 Solve system of linear equations, using matrix method.
5x + 2y = 4
7x + 3y = 5
Answer. The given system of equations can be written in the form of AX = B, where
Page 52
A= [ ] []
5
7
2
3
,X =
x
y
and B = [] 4
5
Now, | A | = 15 − 14 = 1 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
1
A − 1 = | A | (adjA)
∴ A −1 = [ 3
−7
−2
5 ]
∴ X = A − 1B = [ −7
3 −2
5][ ] 4
5
⇒
[] [
x
y
=
12 − 10
− 28 + 25 ] [ ]
=
−3
2
Hence, x=2 and y=-3
Page : 136 , Block Name : Exercise 4.6
Q8 Solve system of linear equations, using matrix method.
2x − y = − 2
3x + 4y = 3
Answer. The given system of equations can be written in the form of AX = B, where
A= [ ] []
2
3
−1
4
,X =
x
y
and B = [ ] −2
3
| A | = 8 + 3 = 11 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now,
1
A − 1 = | A | adj A = 11
[ ]1 4
−3
1
2
∴ X = A − 1B = 11
1
[ ][ ]
−3
4 1
2
−2
3
[]
5
− 11
⇒
[] [
x
y
= 11
1 −8 + 3
6+6 ] [ ]
= 11
1 −5
12
= 12
11
−5 12
Hence, x = 11 and y = 11
Page : 136 , Block Name : Exercise 4.6
5
7
2
3
,X =
x
y
and B = [] 4
5
Now, | A | = 15 − 14 = 1 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
1
A − 1 = | A | (adjA)
∴ A −1 = [ 3
−7
−2
5 ]
∴ X = A − 1B = [ −7
3 −2
5][ ] 4
5
⇒
[] [
x
y
=
12 − 10
− 28 + 25 ] [ ]
=
−3
2
Hence, x=2 and y=-3
Page : 136 , Block Name : Exercise 4.6
Q8 Solve system of linear equations, using matrix method.
2x − y = − 2
3x + 4y = 3
Answer. The given system of equations can be written in the form of AX = B, where
A= [ ] []
2
3
−1
4
,X =
x
y
and B = [ ] −2
3
| A | = 8 + 3 = 11 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now,
1
A − 1 = | A | adj A = 11
[ ]1 4
−3
1
2
∴ X = A − 1B = 11
1
[ ][ ]
−3
4 1
2
−2
3
[]
5
− 11
⇒
[] [
x
y
= 11
1 −8 + 3
6+6 ] [ ]
= 11
1 −5
12
= 12
11
−5 12
Hence, x = 11 and y = 11
Page : 136 , Block Name : Exercise 4.6
Page 53
Q9 Solve system of linear equations, using matrix method.
4x − 3y = 3
3x − 5y = 7
Answer. The given system of equations can be written in the form of AX = B, where
A=
[ ] []
4
3
−3
−5
,X =
x
y
and B =
[] 3
7
| A | = − 20 + 9 = − 11 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now ,
1
A − 1 = | A | (adjA) = − 11 [ ] [ ] 1 −5
−3
3
4
1
= 11
5
3
−3
−4
∴ X = A − 1B = 11
1
[ ][ ]
5
3
−3
−4
3
7
[]
6
− 11
⇒
[ ] [ ][ ] [
x
y
= 11
1 5
3
−3
−4
3
7
= 11
1 15 − 21
9 − 28 ] [ ] [ ]
1
= 11
−6
9 − 28
1
= 11
−6
− 19
= 19
− 11
−6 − 19
Hence,x = 11 and y = 11
Page : 136 , Block Name : Exercise 4.6
Q10 Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
Answer. The given system of equations can be written in the form of AX = B, where
A=
[ ] []
5
3
2
2
,X =
x
y
and B =
[] 3
5
.
| A | = 10 − 6 = 4 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Page : 136 , Block Name : Exercise 4.6
Q11 Solve system of linear equations, using matrix method.
4x − 3y = 3
3x − 5y = 7
Answer. The given system of equations can be written in the form of AX = B, where
A=
[ ] []
4
3
−3
−5
,X =
x
y
and B =
[] 3
7
| A | = − 20 + 9 = − 11 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now ,
1
A − 1 = | A | (adjA) = − 11 [ ] [ ] 1 −5
−3
3
4
1
= 11
5
3
−3
−4
∴ X = A − 1B = 11
1
[ ][ ]
5
3
−3
−4
3
7
[]
6
− 11
⇒
[ ] [ ][ ] [
x
y
= 11
1 5
3
−3
−4
3
7
= 11
1 15 − 21
9 − 28 ] [ ] [ ]
1
= 11
−6
9 − 28
1
= 11
−6
− 19
= 19
− 11
−6 − 19
Hence,x = 11 and y = 11
Page : 136 , Block Name : Exercise 4.6
Q10 Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
Answer. The given system of equations can be written in the form of AX = B, where
A=
[ ] []
5
3
2
2
,X =
x
y
and B =
[] 3
5
.
| A | = 10 − 6 = 4 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Page : 136 , Block Name : Exercise 4.6
Q11 Solve system of linear equations, using matrix method.
Page 54
2x + y + z = 1
3
x − 2y − z = 2
3y − 5z = 9
Answer. The given system of equations can be written in the form of AX = B, where
[ ] [] []
1
2 1 1 x
3
A= 1 −2 −1 , X = y and B = 2
0 3 −5 z 9
| A | = 2(10 + 3) − 1( − 5 − 3) + 0 = 2(13) − 1( − 8) = 26 + 8 = 34 ≠ 0
Now , A 11 = 13, A 12 = 5, A 13 = 3
A 21 = 8, A 22 = − 10, A 23 = − 6
A 31 = 1, A 32 = 3, A 33 = − 5
[ ]
13 8 1
1 1
∴ A − 1 = | A | (adjA) = 34 5 − 10 3
3 −6 −5
[ ][ ]
1
13 8 1
1 3
∴ X = A − 1B = 34 5 − 10 3
2
3 −6 −5
9
[] [ ]
x 13 + 12 + 9
1
⇒ y = 34 5 − 15 + 27
z 3 − 9 − 45
[]
1
[ ]
34 1
1
= 34 17 = 2 Hence,
− 51 3
−2
1 3
x = 1, y = 2 , and z = − 2
Page : 136 , Block Name : Exercise 4.6
Q12 Solve system of linear equations, using matrix method.
3
x − 2y − z = 2
3y − 5z = 9
Answer. The given system of equations can be written in the form of AX = B, where
[ ] [] []
1
2 1 1 x
3
A= 1 −2 −1 , X = y and B = 2
0 3 −5 z 9
| A | = 2(10 + 3) − 1( − 5 − 3) + 0 = 2(13) − 1( − 8) = 26 + 8 = 34 ≠ 0
Now , A 11 = 13, A 12 = 5, A 13 = 3
A 21 = 8, A 22 = − 10, A 23 = − 6
A 31 = 1, A 32 = 3, A 33 = − 5
[ ]
13 8 1
1 1
∴ A − 1 = | A | (adjA) = 34 5 − 10 3
3 −6 −5
[ ][ ]
1
13 8 1
1 3
∴ X = A − 1B = 34 5 − 10 3
2
3 −6 −5
9
[] [ ]
x 13 + 12 + 9
1
⇒ y = 34 5 − 15 + 27
z 3 − 9 − 45
[]
1
[ ]
34 1
1
= 34 17 = 2 Hence,
− 51 3
−2
1 3
x = 1, y = 2 , and z = − 2
Page : 136 , Block Name : Exercise 4.6
Q12 Solve system of linear equations, using matrix method.
Page 55
x−y+z=4
2x + y − 3z = 0
x+y+z=2
Answer. The given system of equations can be written in the form of AX = B, where
[ ] [] []
1 −1 1 x 4
A= 2 1 −3 , X = y and B = 0
1 1 1 z 2
| A | = 1(1 + 3) + 1(2 + 3) + 1(2 − 1) = 4 + 5 + 1 = 10 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now , A 11 = 4, A 12 = − 5, A 13 = 1
A 21 = 2, A 22 = 0, A 23 = − 2
A 31 = 2, A 32 = 5, A 33 = 3
[ ]
4 2 2
1 1
∴ A − 1 = | A | (adjA) = 10 − 5 0 5
1 −2 3
[ ][ ]
4 2 2 0
1
∴ X = A − 1B = 10 − 5 0 5 0
1 −2 3 2
[] [ ]
x 16 + 0 + 4
1
[]
⇒ y = − 20 + 0 + 10
10 2
y 4+0+6
= −1
[ ]
1
− 10
1
= − 10 Hence, x = 2, y = − 1, and z = 1
10
10
Page : 136 , Block Name : Exercise 4.6
Q13 Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x − 2y + z = − 4
3x − y − 2z = 3
Answer. The given system of equations can be written in the form AX = B, where
2x + y − 3z = 0
x+y+z=2
Answer. The given system of equations can be written in the form of AX = B, where
[ ] [] []
1 −1 1 x 4
A= 2 1 −3 , X = y and B = 0
1 1 1 z 2
| A | = 1(1 + 3) + 1(2 + 3) + 1(2 − 1) = 4 + 5 + 1 = 10 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now , A 11 = 4, A 12 = − 5, A 13 = 1
A 21 = 2, A 22 = 0, A 23 = − 2
A 31 = 2, A 32 = 5, A 33 = 3
[ ]
4 2 2
1 1
∴ A − 1 = | A | (adjA) = 10 − 5 0 5
1 −2 3
[ ][ ]
4 2 2 0
1
∴ X = A − 1B = 10 − 5 0 5 0
1 −2 3 2
[] [ ]
x 16 + 0 + 4
1
[]
⇒ y = − 20 + 0 + 10
10 2
y 4+0+6
= −1
[ ]
1
− 10
1
= − 10 Hence, x = 2, y = − 1, and z = 1
10
10
Page : 136 , Block Name : Exercise 4.6
Q13 Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x − 2y + z = − 4
3x − y − 2z = 3
Answer. The given system of equations can be written in the form AX = B, where
Page 56
[ ] [] [ ]
2 3 3 x 5
A= 1 −2 1 ,X = y and B = −4
3 −2 −2 z 3
| A | = 2(4 + 1) − 3( − 2 − 3) + 3( − 1 + 6) = 2(5) − 3( − 5) + 3(5) = 10 + 15 + 15 = 40 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now, A11 = 5, A 12 = 5, A 13 = 5
A 21 = 3, A 22 = − 13, A 23 = 11
A 31 = 9, A 32 = 1, A 33 = − 7
[ ]
5 − 13 1
1 1
∴ A −1 = (adjA) = 5 − 13 1
|A| 40
5 11 −7
[ ][ ]
5 3 9
5
1 5 − 13 1
∴ X = A − 1B = 40 −4
5 − 13 1
3
5 11 −7
[] [ ]
x 25 + 52 + 3
1
⇒ y = 40 25 + 52 + 3
z 25 − 44 − 21
[ ]
40
1
= 80
40
− 40
[]
1
= 2
−1
Hence , x = 1, y = 2, and z = − 1
Page : 136 , Block Name : Exercise 4.6
Q14 Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = − 5
2x − y + 3z = 12
Answer. The given system of equations can be written in the form of AX = B, where
2 3 3 x 5
A= 1 −2 1 ,X = y and B = −4
3 −2 −2 z 3
| A | = 2(4 + 1) − 3( − 2 − 3) + 3( − 1 + 6) = 2(5) − 3( − 5) + 3(5) = 10 + 15 + 15 = 40 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now, A11 = 5, A 12 = 5, A 13 = 5
A 21 = 3, A 22 = − 13, A 23 = 11
A 31 = 9, A 32 = 1, A 33 = − 7
[ ]
5 − 13 1
1 1
∴ A −1 = (adjA) = 5 − 13 1
|A| 40
5 11 −7
[ ][ ]
5 3 9
5
1 5 − 13 1
∴ X = A − 1B = 40 −4
5 − 13 1
3
5 11 −7
[] [ ]
x 25 + 52 + 3
1
⇒ y = 40 25 + 52 + 3
z 25 − 44 − 21
[ ]
40
1
= 80
40
− 40
[]
1
= 2
−1
Hence , x = 1, y = 2, and z = − 1
Page : 136 , Block Name : Exercise 4.6
Q14 Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = − 5
2x − y + 3z = 12
Answer. The given system of equations can be written in the form of AX = B, where
Page 57
[ ] [] [ ]
1 −1 2 x 7
A= 3 4 −5 , X = y and B = −5
2 −1 3 z 12
now, | A | = 1(12 − 5) + 1(9 + 10) + 2( − 3 − 8) = 7 + 19 − 22 = 4 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now, A 11 = 7, A 12 = − 19, A 13 = − 11
A 21 = 1, A 22 = − 1, A 23 = − 1
A 31 = − 3, A 32 = 11, A 33 = 7
1
∴ A − 1 = | A | (adjA) = 4
1
[ − 11
7 1
−1
11
7 ]
[ ][ ]
7 1 −3 7
1
∴ X = A − 1B = 4 − 19 −1 11 −5
− 11 −1 7 12
[] [ ]
x 49 − 5 − 36
1
⇒ y = − 133 + 5 + 132
4
y − 77 + 5 + 84
[] []
8
1 1
= 4 =
4 3
4
Hence, x = 2, y = 1, and z = 3
Page : 136 , Block Name : Exercise 4.6
[ ]
2 −3 5 2x − 3y + 5z = 11
Q15 If A = 3 2 −4 , find A − 1. Using A − 1 solve the system of equations 3x + 2y − 4z = −5
1 1 −2 x + y − 2z = − 3
Answer.
1 −1 2 x 7
A= 3 4 −5 , X = y and B = −5
2 −1 3 z 12
now, | A | = 1(12 − 5) + 1(9 + 10) + 2( − 3 − 8) = 7 + 19 − 22 = 4 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now, A 11 = 7, A 12 = − 19, A 13 = − 11
A 21 = 1, A 22 = − 1, A 23 = − 1
A 31 = − 3, A 32 = 11, A 33 = 7
1
∴ A − 1 = | A | (adjA) = 4
1
[ − 11
7 1
−1
11
7 ]
[ ][ ]
7 1 −3 7
1
∴ X = A − 1B = 4 − 19 −1 11 −5
− 11 −1 7 12
[] [ ]
x 49 − 5 − 36
1
⇒ y = − 133 + 5 + 132
4
y − 77 + 5 + 84
[] []
8
1 1
= 4 =
4 3
4
Hence, x = 2, y = 1, and z = 3
Page : 136 , Block Name : Exercise 4.6
[ ]
2 −3 5 2x − 3y + 5z = 11
Q15 If A = 3 2 −4 , find A − 1. Using A − 1 solve the system of equations 3x + 2y − 4z = −5
1 1 −2 x + y − 2z = − 3
Answer.
Page 58
[ ]
2 −3 5
A= 3 2 −4
1 1 −2
∴ | A | = 2( − 4 + 4) + 3( − 6 + 4) + 5(3 − 2) = 0 − 6 + 5 = − 1 ≠ 0
Now, A 11 = 0, A 12 = 2, A 13 = 1
A 21 = − 1, A 22 = − 9, A 23 = − 5
A 31 = 2, A 32 = 23, A 33 = 13
[ ][ ]
0 −1 2 0 1 −2
1
∴ A − 1 = | A | (adjA) = − 2 −9 23 = −2 9 − 23 ....(i)
1 −5 13 −1 5 − 13
Now, the given system of equations can be written in the form of AX = B, where
[ ] [] [ ]
2 −3 5 x 11
A= 3 2 −4 , X = y and B = −5
1 1 −2 z −3
The solution of the system of equations is given by X = A − 1B
X = A − 1B
[] [ ][ ]
x 0 1 −2 11
⇒ y = −2 9 − 23 −5 [ Using (i)]
z −1 5 − 13 −3
[ ]
0−5+6
= − 22 − 45 + 69
− 11 − 25 + 39
[]
1
= 2
3
Hence, x = 1, y = 2, and z = 3
Page : 137 , Block Name : Exercise 4.6
Q16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg
wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.
Find cost of each item per kg by matrix method.
Answer. Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.
Then, the given situation can be represented by a system of equations as:
2 −3 5
A= 3 2 −4
1 1 −2
∴ | A | = 2( − 4 + 4) + 3( − 6 + 4) + 5(3 − 2) = 0 − 6 + 5 = − 1 ≠ 0
Now, A 11 = 0, A 12 = 2, A 13 = 1
A 21 = − 1, A 22 = − 9, A 23 = − 5
A 31 = 2, A 32 = 23, A 33 = 13
[ ][ ]
0 −1 2 0 1 −2
1
∴ A − 1 = | A | (adjA) = − 2 −9 23 = −2 9 − 23 ....(i)
1 −5 13 −1 5 − 13
Now, the given system of equations can be written in the form of AX = B, where
[ ] [] [ ]
2 −3 5 x 11
A= 3 2 −4 , X = y and B = −5
1 1 −2 z −3
The solution of the system of equations is given by X = A − 1B
X = A − 1B
[] [ ][ ]
x 0 1 −2 11
⇒ y = −2 9 − 23 −5 [ Using (i)]
z −1 5 − 13 −3
[ ]
0−5+6
= − 22 − 45 + 69
− 11 − 25 + 39
[]
1
= 2
3
Hence, x = 1, y = 2, and z = 3
Page : 137 , Block Name : Exercise 4.6
Q16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg
wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.
Find cost of each item per kg by matrix method.
Answer. Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.
Then, the given situation can be represented by a system of equations as:
Page 59
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
This system of equations can be written in the form of AX = B, where
[ ] []
4 3 2
60
A=
2
6
4
4
6
6
,X =
[]x
z
and B = 90
70
6 2 3
| A | = 4(12 − 12) − 3(6 − 36) + 2(4 − 24) = 0 + 90 − 40 = 50 ≠ 0
Now, A 11 = 0, A 12 = 30, A 13 = − 20
A 21 = − 5, A 22 = 0, A 23 = 10
A 31 = 10, A 32 = − 20, A 33 = 10
[ ]
0 −5 10
∴ adjA = 30 0 − 20
− 20 10 10
[ ]
now ,
0 −5 10
1 1 0 0 − 20
∴ A − 1 = | A | adjA = 50
30 10 − 20
− 20 10 10
[ ][ ]
0 −5 10 60
1
⇒ X = 50 30 0 − 20 90
− 20 10 10 70
[] [ ]
x 0 − 450 + 700
1
⇒ y = 50 1800 + 0 − 1400
z − 1200 + 900 + 700
[]
250
1
= 400
50
400
[]
5
= 8
8
∴ x = 5, y = 8, and z = 8
Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8
per kg.
2x + 4y + 6z = 90
6x + 2y + 3z = 70
This system of equations can be written in the form of AX = B, where
[ ] []
4 3 2
60
A=
2
6
4
4
6
6
,X =
[]x
z
and B = 90
70
6 2 3
| A | = 4(12 − 12) − 3(6 − 36) + 2(4 − 24) = 0 + 90 − 40 = 50 ≠ 0
Now, A 11 = 0, A 12 = 30, A 13 = − 20
A 21 = − 5, A 22 = 0, A 23 = 10
A 31 = 10, A 32 = − 20, A 33 = 10
[ ]
0 −5 10
∴ adjA = 30 0 − 20
− 20 10 10
[ ]
now ,
0 −5 10
1 1 0 0 − 20
∴ A − 1 = | A | adjA = 50
30 10 − 20
− 20 10 10
[ ][ ]
0 −5 10 60
1
⇒ X = 50 30 0 − 20 90
− 20 10 10 70
[] [ ]
x 0 − 450 + 700
1
⇒ y = 50 1800 + 0 − 1400
z − 1200 + 900 + 700
[]
250
1
= 400
50
400
[]
5
= 8
8
∴ x = 5, y = 8, and z = 8
Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8
per kg.
Page 60
Page : 137 , Block Name : Exercise 4.6
| |
x sinθ cosθ
Q1 Prove that the determinant − sinθ −x 1
cosθ 1 x is independent of θ
Answer.
| |
x sinθ cosθ
Δ = − sinθ −x 1
cosθ 1 x
( )
= x x 2 − 1 − sinθ( − xsinθ − cosθ) + cosθ( − sinθ + xcosθ)
= x 3 − x + xsin 2θ + sinθcosθ − sinθcosθ + xcos 2θ
(
= x 3 − x + x sin 2θ + cos 2θ )
= x3 − x + x
= x 3( Independent of θ)
Hence , Δ is independent of θ
Page : 141 , Block Name : Miscellaneous Exercise
Q2 Without expanding the determinant, prove that
| || |
a a2 bc 1 a2 a3
b b2 ca = 1 b2 b3
c c2 ab 1 c2 c3
| |
a a2 bc
Answer. L.H.S. = b b 2 ca
c c2 ab
| |
a2 a3 abc
1
= abc b 2 b3 abc [R1 → aR1, R2 → bR2, and R3 → cR3 ]
c2 c3 abc
| |
x sinθ cosθ
Q1 Prove that the determinant − sinθ −x 1
cosθ 1 x is independent of θ
Answer.
| |
x sinθ cosθ
Δ = − sinθ −x 1
cosθ 1 x
( )
= x x 2 − 1 − sinθ( − xsinθ − cosθ) + cosθ( − sinθ + xcosθ)
= x 3 − x + xsin 2θ + sinθcosθ − sinθcosθ + xcos 2θ
(
= x 3 − x + x sin 2θ + cos 2θ )
= x3 − x + x
= x 3( Independent of θ)
Hence , Δ is independent of θ
Page : 141 , Block Name : Miscellaneous Exercise
Q2 Without expanding the determinant, prove that
| || |
a a2 bc 1 a2 a3
b b2 ca = 1 b2 b3
c c2 ab 1 c2 c3
| |
a a2 bc
Answer. L.H.S. = b b 2 ca
c c2 ab
| |
a2 a3 abc
1
= abc b 2 b3 abc [R1 → aR1, R2 → bR2, and R3 → cR3 ]
c2 c3 abc
Page 61
| |
a2 a3 1
1
= abc ⋅ abc b 2 b3 1 [ Taking out factor abc from C3 ]
c2 c3 1
| |
a2 a3 1
= b2 b3 1
c2 c3 1
| |
1 a2 a3
= 1 b2 b3 [Applying C 1 ↔ C 3 and C 2 ↔ C 3]
1 c2 c3
=R.H.S.
Page : 141 , Block Name : Miscellaneous Exercise
| |
cosαcosβ cosαsinβ − sinα
Q3 Evaluate − sinβ cosβ 0
sinαcosβ sinαsinβ cosα
Answer.
| |
cosαcosβ cosαsinβ − sinα
Δ= − sinβ cosβ 0
sinαcosβ sinαsinβ cosα
Expanding along C 3, we have:
( )
Δ = − sinα − sinαsin 2β − cos 2βsinα + cosα cosαcos 2β + cosαsin 2β ( )
( )
= sin 2α sin 2β + cos 2β + cos 2α cos 2β + sin 2β ( )
= sin 2α(1) + cos 2α(1)
=1
Page : 141 , Block Name : Miscellaneous Exercise
Q4 If a, b and c are real numbers, and
a2 a3 1
1
= abc ⋅ abc b 2 b3 1 [ Taking out factor abc from C3 ]
c2 c3 1
| |
a2 a3 1
= b2 b3 1
c2 c3 1
| |
1 a2 a3
= 1 b2 b3 [Applying C 1 ↔ C 3 and C 2 ↔ C 3]
1 c2 c3
=R.H.S.
Page : 141 , Block Name : Miscellaneous Exercise
| |
cosαcosβ cosαsinβ − sinα
Q3 Evaluate − sinβ cosβ 0
sinαcosβ sinαsinβ cosα
Answer.
| |
cosαcosβ cosαsinβ − sinα
Δ= − sinβ cosβ 0
sinαcosβ sinαsinβ cosα
Expanding along C 3, we have:
( )
Δ = − sinα − sinαsin 2β − cos 2βsinα + cosα cosαcos 2β + cosαsin 2β ( )
( )
= sin 2α sin 2β + cos 2β + cos 2α cos 2β + sin 2β ( )
= sin 2α(1) + cos 2α(1)
=1
Page : 141 , Block Name : Miscellaneous Exercise
Q4 If a, b and c are real numbers, and
Page 62
| |
b+c c+a a+b
Δ= c+a a+b b+c =0
a+b b+c c+a
Show that either a + b + c = 0 or a = b = c.
Answer.
| |
b+c c+a a+b
Δ= c+a a+b b+c =0
a+b b+c c+a
Applying R 1 → R 1 + R 2 + R 3,
| |
2(a + b + c) 2(a + b + c) 2(a + b + c)
Δ= c+a a+b b+c
a+b b+c c+a
| |
1 1 1
= 2(a + b + c) c + a a+b b+c
a+b b+c c+a
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = 2(a + b + c) c + a b−c b−a
a+b c−a c−b
Expanding along R 1, we have:
Δ = 2(a + b + c)(1)[(b − c)(c − b) − (b − a)(c − a)]
[ ]
= 2(a + b + c) − b 2 − c 2 + 2bc − bc + ba + ac − a 2
= 2(a + b + c) [ab + bc + ca − a − b − c ] 2 2 2
It is given that Δ = 0
[
(a + b + c) ab + bc + ca − a 2 − b 2 − c 2 = 0 ]
⇒ Either a + b + c = 0, or ab + bc + ca − a − b − c 2 = 0 2 2
Now,
ab + bc + ca − a 2 − b 2 − c 2 = 0
⇒ − 2ab − 2bc − 2ca + 2a 2 + 2b 2 + 2c 2 = 0
⇒ (a − b) 2 + (b − c) 2 + (c − a) 2 = 0
⇒ (a − b) 2 = (b − c) 2 = (c − a) 2 = 0 [(a − b) , (b − c) , (c − a) are non-negative ]
2 2 2
b+c c+a a+b
Δ= c+a a+b b+c =0
a+b b+c c+a
Show that either a + b + c = 0 or a = b = c.
Answer.
| |
b+c c+a a+b
Δ= c+a a+b b+c =0
a+b b+c c+a
Applying R 1 → R 1 + R 2 + R 3,
| |
2(a + b + c) 2(a + b + c) 2(a + b + c)
Δ= c+a a+b b+c
a+b b+c c+a
| |
1 1 1
= 2(a + b + c) c + a a+b b+c
a+b b+c c+a
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = 2(a + b + c) c + a b−c b−a
a+b c−a c−b
Expanding along R 1, we have:
Δ = 2(a + b + c)(1)[(b − c)(c − b) − (b − a)(c − a)]
[ ]
= 2(a + b + c) − b 2 − c 2 + 2bc − bc + ba + ac − a 2
= 2(a + b + c) [ab + bc + ca − a − b − c ] 2 2 2
It is given that Δ = 0
[
(a + b + c) ab + bc + ca − a 2 − b 2 − c 2 = 0 ]
⇒ Either a + b + c = 0, or ab + bc + ca − a − b − c 2 = 0 2 2
Now,
ab + bc + ca − a 2 − b 2 − c 2 = 0
⇒ − 2ab − 2bc − 2ca + 2a 2 + 2b 2 + 2c 2 = 0
⇒ (a − b) 2 + (b − c) 2 + (c − a) 2 = 0
⇒ (a − b) 2 = (b − c) 2 = (c − a) 2 = 0 [(a − b) , (b − c) , (c − a) are non-negative ]
2 2 2
Page 63
⇒ (a − b) = (b − c) = (c − a) = 0
⇒a=b=c
Hence, if Δ = 0, then either a + b + c = 0 or a = b = c .
Page : 141 , Block Name : Miscellaneous Exercise
Q5 Solve the equations :
| |
x+a x x
x x+a x = 0, a ≠ 0
x x x+a
Answer.
| |
x+a x x
x x+a x =0
x x x+a
Applying R 1 → R 1 + R 2 + R 3, , we get :
| |
3x + a 3x + a 3x + a
x x+a x =0
x x x+a
| |
1 1 1
⇒ (3x + a) x x+a x =0
x x x+a
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, , we have :
| |
1 0 0
(3x + a) x a 0 =0
x 0 a
Expanding along R 1, we have:
[
(3x + a) 1 × a 2 = 0 ]
⇒ a 2(3x + a) = 0
But a ≠ 0
Therefore, we have:
3x + a = 0
a
⇒x= − 3
Page : 141 , Block Name : Miscellaneous Exercise
⇒a=b=c
Hence, if Δ = 0, then either a + b + c = 0 or a = b = c .
Page : 141 , Block Name : Miscellaneous Exercise
Q5 Solve the equations :
| |
x+a x x
x x+a x = 0, a ≠ 0
x x x+a
Answer.
| |
x+a x x
x x+a x =0
x x x+a
Applying R 1 → R 1 + R 2 + R 3, , we get :
| |
3x + a 3x + a 3x + a
x x+a x =0
x x x+a
| |
1 1 1
⇒ (3x + a) x x+a x =0
x x x+a
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, , we have :
| |
1 0 0
(3x + a) x a 0 =0
x 0 a
Expanding along R 1, we have:
[
(3x + a) 1 × a 2 = 0 ]
⇒ a 2(3x + a) = 0
But a ≠ 0
Therefore, we have:
3x + a = 0
a
⇒x= − 3
Page : 141 , Block Name : Miscellaneous Exercise
Page 64
| |
a2 bc ac + c 2
Q6 Prove that a 2 + ab b2 ac = 4a 2b 2c 2
ab b 2 + bc c2
Answer.
| |
a2 bc ac + c 2
Δ = a 2 + ab b2 ac
ab b 2 + bc c2
Taking out common factors a, b, and c from C 1, C 2, and C 3, we have:
| |
a c a+c
Δ = abc a + b b a
b b+c c
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
| |
a c a+c
Δ = abc b b−c −c
b−a b −a
Applying R 2 → R 2 + R 1 , we have :
| |
a c a+c
Δ = abc a + b b a
b−a b −a
Applying R 3 → R 3 + R 2
| |
a c a+c
Δ = abc a + b b a
2b 2b 0
| |
a c a+c
= 2ab 2c a c a+c
1 1 0
Applying C 2 → C 2 − C 1
a2 bc ac + c 2
Q6 Prove that a 2 + ab b2 ac = 4a 2b 2c 2
ab b 2 + bc c2
Answer.
| |
a2 bc ac + c 2
Δ = a 2 + ab b2 ac
ab b 2 + bc c2
Taking out common factors a, b, and c from C 1, C 2, and C 3, we have:
| |
a c a+c
Δ = abc a + b b a
b b+c c
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
| |
a c a+c
Δ = abc b b−c −c
b−a b −a
Applying R 2 → R 2 + R 1 , we have :
| |
a c a+c
Δ = abc a + b b a
b−a b −a
Applying R 3 → R 3 + R 2
| |
a c a+c
Δ = abc a + b b a
2b 2b 0
| |
a c a+c
= 2ab 2c a c a+c
1 1 0
Applying C 2 → C 2 − C 1
Page 65
| |
a c−a a+c
Δ = 2ab 2c a + b −a a
1 0 0
Δ = 2ab 2c[a(c − a) + a(a + c)]
= 2ab 2c[a(c − a) + a(a + c)]
[
= 2ab 2c ac − a 2 + a 2 + ac ]
= 2ab 2c(2ac)
= 4a 2b 2c 2
Hence, the given result is proved.
Page : 141 , Block Name : Miscellaneous Exercise
[ ] [ ]
3 −1 1 1 2 −2
Q7 If A − 1 = − 15 6 − 5 and B = −1 3 0 , find (AB) +
5 −2 2 0 −2 1
Answer.
We know that (AB) − 1 = B − 1A − 1
[ ]
1 2 −2
B= −1 3 0
0 −2 1
∴ | B | = 1 × 3 − 2 × ( − 1) − 2(2) = 3 + 2 − 4 = 5 − 4 = 1
Now, A 11 = 3, A 12 = 1, A 13 = 2
A 21 = 2, A 22 = 1, A 23 = 2
A 31 = 6, A 32 = 2, A 33 = 5
[ ]
3 2 6
∴ adjB = 1 1 2
2 2 5
1
B − 1 = | B | ⋅ adjB
[ ]
3 2 6
∴ B −1 = 1 1 2
2 2 5
∴ (AB) − 1 = B − 1A − 1
a c−a a+c
Δ = 2ab 2c a + b −a a
1 0 0
Δ = 2ab 2c[a(c − a) + a(a + c)]
= 2ab 2c[a(c − a) + a(a + c)]
[
= 2ab 2c ac − a 2 + a 2 + ac ]
= 2ab 2c(2ac)
= 4a 2b 2c 2
Hence, the given result is proved.
Page : 141 , Block Name : Miscellaneous Exercise
[ ] [ ]
3 −1 1 1 2 −2
Q7 If A − 1 = − 15 6 − 5 and B = −1 3 0 , find (AB) +
5 −2 2 0 −2 1
Answer.
We know that (AB) − 1 = B − 1A − 1
[ ]
1 2 −2
B= −1 3 0
0 −2 1
∴ | B | = 1 × 3 − 2 × ( − 1) − 2(2) = 3 + 2 − 4 = 5 − 4 = 1
Now, A 11 = 3, A 12 = 1, A 13 = 2
A 21 = 2, A 22 = 1, A 23 = 2
A 31 = 6, A 32 = 2, A 33 = 5
[ ]
3 2 6
∴ adjB = 1 1 2
2 2 5
1
B − 1 = | B | ⋅ adjB
[ ]
3 2 6
∴ B −1 = 1 1 2
2 2 5
∴ (AB) − 1 = B − 1A − 1
Page 66
[ ][ ]
3 2 6 3 −1 1
= 1 1 2 − 15 6 −5
2 2 5 5 −2 2
[ ]
9 − 30 + 30 − 3 + 12 − 12 3 − 10 + 12
= 3 − 15 + 10 −1 + 6 − 4 1−5+4
6 − 30 + 25 − 2 + 12 − 10 2 − 10 + 10
[ ]
9 −3 5
= −2 1 0
1 0 2
Page : 141 , Block Name : Miscellaneous Exercise
[ ]
1 −2 1
Q8 Let A = −2 3 1
1 1 5 verify that
(i)[adjA] − 1 = adj A − 1 ( )
( ) =A
(ii) A − 1
−1
Answer.
[ ]
1 −2 1
A= −2 3 1
1 1 5
∴ | A | = 1(15 − 1) + 2( − 10 − 1) + 1( − 2 − 3) = 14 − 22 − 5 = − 13
Now, A 11 = 14, A 12 = 11, A 13 = − 5
A 21 = 11, A 22 = 4, A 23 = − 3
A 31 = − 5, A 12 = − 3, A 13 = − 1
[ ]
14 11 −5
∴ adjA = 11 4 −3
−5 −3 −1
3 2 6 3 −1 1
= 1 1 2 − 15 6 −5
2 2 5 5 −2 2
[ ]
9 − 30 + 30 − 3 + 12 − 12 3 − 10 + 12
= 3 − 15 + 10 −1 + 6 − 4 1−5+4
6 − 30 + 25 − 2 + 12 − 10 2 − 10 + 10
[ ]
9 −3 5
= −2 1 0
1 0 2
Page : 141 , Block Name : Miscellaneous Exercise
[ ]
1 −2 1
Q8 Let A = −2 3 1
1 1 5 verify that
(i)[adjA] − 1 = adj A − 1 ( )
( ) =A
(ii) A − 1
−1
Answer.
[ ]
1 −2 1
A= −2 3 1
1 1 5
∴ | A | = 1(15 − 1) + 2( − 10 − 1) + 1( − 2 − 3) = 14 − 22 − 5 = − 13
Now, A 11 = 14, A 12 = 11, A 13 = − 5
A 21 = 11, A 22 = 4, A 23 = − 3
A 31 = − 5, A 12 = − 3, A 13 = − 1
[ ]
14 11 −5
∴ adjA = 11 4 −3
−5 −3 −1
Page 67
1
∴ A −1 = (adjA)
|A|
[ ] [ ]
14 11 −5 − 14 − 11 5
1 1
= − 11 4 −3 = − 11 −4 3
13 13
−5 −3 −1 5 3 1
(i)
| adjA | = 14( − 4 − 9) − 11( − 11 − 15) − 5( − 33 + 20)
= 14( − 13) − 11( − 26) − 5( − 13)
= − 182 + 286 + 65 = 169
We have,
[ ]
− 13 26 − 13
adj(adjA) = 26 − 39 − 13
− 13 − 13 − 65
1
∴ [adjA] 1 = (adj(adjA))
| adjA |
[ ]
− 13 26 − 13
1
= − 13 − 39 − 13
169
− 13 − 13 − 65
[ ]
−1 2 −1
1
= 13 2 −3 −1
−1 −1 −5
[ ]
14 11 5
− 13 − 13
[ ]
13
− 14 − 11 5
1 11 4 3
A − 1 = 13 − 11 −4 3 = − 13 − 13 13
5 3 1 5 3 1
13 13 13
[ ]
4
− 169 − 169
9
(
− − 169 − 169
11 15
) 33 20
− 169 + 169
( )
∴ adj A − 1 =
(
− − 169 − 169
11 15
) 14
− 169 − 169
25
33
− 169 + 169
20
(
− − 169 + 169
42 55
) 56
169
121
− 169
∴ A −1 = (adjA)
|A|
[ ] [ ]
14 11 −5 − 14 − 11 5
1 1
= − 11 4 −3 = − 11 −4 3
13 13
−5 −3 −1 5 3 1
(i)
| adjA | = 14( − 4 − 9) − 11( − 11 − 15) − 5( − 33 + 20)
= 14( − 13) − 11( − 26) − 5( − 13)
= − 182 + 286 + 65 = 169
We have,
[ ]
− 13 26 − 13
adj(adjA) = 26 − 39 − 13
− 13 − 13 − 65
1
∴ [adjA] 1 = (adj(adjA))
| adjA |
[ ]
− 13 26 − 13
1
= − 13 − 39 − 13
169
− 13 − 13 − 65
[ ]
−1 2 −1
1
= 13 2 −3 −1
−1 −1 −5
[ ]
14 11 5
− 13 − 13
[ ]
13
− 14 − 11 5
1 11 4 3
A − 1 = 13 − 11 −4 3 = − 13 − 13 13
5 3 1 5 3 1
13 13 13
[ ]
4
− 169 − 169
9
(
− − 169 − 169
11 15
) 33 20
− 169 + 169
( )
∴ adj A − 1 =
(
− − 169 − 169
11 15
) 14
− 169 − 169
25
33
− 169 + 169
20
(
− − 169 + 169
42 55
) 56
169
121
− 169
Page 68
[ ] [ ]
− 13 26 − 13 −1 2 −1
1 1
= 169 26 − 39 − 13 = 13 2 −3 −1
− 13 − 13 − 65 −1 −1 −5
Hence , [adjA] − 1 = adj A − 1 ( )
(ii) We have,
[ ]
− 14 − 11 5
1
A − 1 = 13 − 11 −4 3
5 3 1
[ ]
−1 2 −1
1
And, adjA − 1 = 13 2 −3 −1
−1 −1 −5
|A | = ( ) [ − 14 × ( − 13) + 11 × ( − 26) + 5 × ( − 13)] = ( ) × ( − 169) = −
1 3 1 3 1
−1
13 13 13
( ) [ ][ ]
−1 2 −1 1 −2 1
−1
( ) −1 adjA 1 1
∴ A −1 = = × 13 2 −3 −1 = −2 3 1 =A
|A | −1 1
− 13 −1 −1 −5 1 1 5
( ) =A
∴ A −1
−1
Page : 142 , Block Name : Miscellaneous Exercise
| |
x y x+y
Q9 Evaluate y x+y x
x+y x y
Answer.
| |
x y x+y
Δ= y x+y x
x+y x y
Applying R 1 → R 1 + R 2 + R 3
| |
2(x + y) 2(x + y) 2(x + y)
Δ= y x+y x
x+y x y
− 13 26 − 13 −1 2 −1
1 1
= 169 26 − 39 − 13 = 13 2 −3 −1
− 13 − 13 − 65 −1 −1 −5
Hence , [adjA] − 1 = adj A − 1 ( )
(ii) We have,
[ ]
− 14 − 11 5
1
A − 1 = 13 − 11 −4 3
5 3 1
[ ]
−1 2 −1
1
And, adjA − 1 = 13 2 −3 −1
−1 −1 −5
|A | = ( ) [ − 14 × ( − 13) + 11 × ( − 26) + 5 × ( − 13)] = ( ) × ( − 169) = −
1 3 1 3 1
−1
13 13 13
( ) [ ][ ]
−1 2 −1 1 −2 1
−1
( ) −1 adjA 1 1
∴ A −1 = = × 13 2 −3 −1 = −2 3 1 =A
|A | −1 1
− 13 −1 −1 −5 1 1 5
( ) =A
∴ A −1
−1
Page : 142 , Block Name : Miscellaneous Exercise
| |
x y x+y
Q9 Evaluate y x+y x
x+y x y
Answer.
| |
x y x+y
Δ= y x+y x
x+y x y
Applying R 1 → R 1 + R 2 + R 3
| |
2(x + y) 2(x + y) 2(x + y)
Δ= y x+y x
x+y x y
Page 69
| |
1 x+y x
= 2(x + y) y x+y x
x+y x y
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = 2(x + y) y x x−y
x+y −y −x
Expanding along R1, we have
[
Δ = 2(x + y) − x 2 + y(x − y) ]
= − 2(x + y) x 2 + y(x − y)( ]
(
= − 2 x3 + y3 )
Page : 142 , Block Name : Miscellaneous Exercise
| |
1 x y
Q10 Evaluate 1 x+y y
1 x x+y
Answer.
| |
1 x y
Δ= 1 x+y y
1 x x+y
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
| |
1 x y
Δ= 0 y 0
0 0 x
Expanding along C1, we have
Δ = 1(xy − 0) = xy
Page : 142 , Block Name : Miscellaneous Exercise
Q11 Using properties of determinants, prove that:
1 x+y x
= 2(x + y) y x+y x
x+y x y
Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1, we have:
| |
1 0 0
Δ = 2(x + y) y x x−y
x+y −y −x
Expanding along R1, we have
[
Δ = 2(x + y) − x 2 + y(x − y) ]
= − 2(x + y) x 2 + y(x − y)( ]
(
= − 2 x3 + y3 )
Page : 142 , Block Name : Miscellaneous Exercise
| |
1 x y
Q10 Evaluate 1 x+y y
1 x x+y
Answer.
| |
1 x y
Δ= 1 x+y y
1 x x+y
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
| |
1 x y
Δ= 0 y 0
0 0 x
Expanding along C1, we have
Δ = 1(xy − 0) = xy
Page : 142 , Block Name : Miscellaneous Exercise
Q11 Using properties of determinants, prove that:
Page 70
| |
α α2 β+γ
β β2 γ + α = (β − γ)(γ − α)(α − β)(α + β + γ)
γ γ2 α+β
Answer.
| |
α α2 β+γ
Δ= β β2 γ+α
γ γ2 α+β
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, ,we have
| |
α α2 β+γ
= (β − α)(γ − α) 1 β+α −1
1 γ+α −1
Applying R 3 → R 3 − R 2, we have:
| |
α α2 β+γ
Δ = (β − α)(γ − α) | β+α −1
0 γ−β 0
Expanding along R 3, we have:
Δ = (β − α)(γ − α)[ − (γ − β)( − α − β − γ)]
= (β − α)(γ − α)(γ − β)(α + β + γ)
= (α − β)(β − γ)(γ − α)(α + β + γ)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q12 Using properties of determinants, prove that:
| x2 1 + px 3
|
x
y y2 1 + py 3 = (1 + pxyz)(x − y)(y − z)(z − x)
z z2 1 + pz 3
Answer.
α α2 β+γ
β β2 γ + α = (β − γ)(γ − α)(α − β)(α + β + γ)
γ γ2 α+β
Answer.
| |
α α2 β+γ
Δ= β β2 γ+α
γ γ2 α+β
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, ,we have
| |
α α2 β+γ
= (β − α)(γ − α) 1 β+α −1
1 γ+α −1
Applying R 3 → R 3 − R 2, we have:
| |
α α2 β+γ
Δ = (β − α)(γ − α) | β+α −1
0 γ−β 0
Expanding along R 3, we have:
Δ = (β − α)(γ − α)[ − (γ − β)( − α − β − γ)]
= (β − α)(γ − α)(γ − β)(α + β + γ)
= (α − β)(β − γ)(γ − α)(α + β + γ)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q12 Using properties of determinants, prove that:
| x2 1 + px 3
|
x
y y2 1 + py 3 = (1 + pxyz)(x − y)(y − z)(z − x)
z z2 1 + pz 3
Answer.
Page 71
| |
x x2 1 + px 3
Δ= y y2 1 + py 3
z z2 1 + pz 3
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 3, we have:
x2 1 + px 3
| |
x
Δ=
y−x y2 − x2 ( )
p y3 − x3
z−x z2 − x2 p (z − x )
3 3
x2 1 + px 3
| |
x
= (y − x)(z − x)
1 (
y+x )
p y 2 + x 2 + xy
1 z + x p (z + x + xz ) 2 2
Applying R 3 → R 3 − R 2, we have:
| |
x x2 1 + px 3
Δ = (y − x)(z − x) 1 y+x (
p y 2 + x 2 + xy )
0 z−y p(z − y)(x + y + z)
| |
x x2 1 + px 3
= (y − x)(z − x)(z − y) 1 y+x (
p y 2 + x 2 + xy )
0 1 p(x + y + z)
Expanding along R 3, we have:
[ ( )
Δ = (x − y)(y − z)(z − x) ( − 1)(p) xy 2 + x 3 + x 2y + 1 + px 3 + p(x + y + z)(xy)
= (x − y)(y − z)(z − x) [ − pxy − px − px y + 1 + px + px y + pxy + pxyz ]
2 3 2 3 2 2
= (x − y)(y − z)(z − x)(1 + pxyz)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q13 Using properties of determinants, prove that:
| |
3a −a + b −a + c
−b + a 3b − b + c = 3(a + b + c)(ab + bc + ca)
−c + a −c + b 3c
x x2 1 + px 3
Δ= y y2 1 + py 3
z z2 1 + pz 3
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 3, we have:
x2 1 + px 3
| |
x
Δ=
y−x y2 − x2 ( )
p y3 − x3
z−x z2 − x2 p (z − x )
3 3
x2 1 + px 3
| |
x
= (y − x)(z − x)
1 (
y+x )
p y 2 + x 2 + xy
1 z + x p (z + x + xz ) 2 2
Applying R 3 → R 3 − R 2, we have:
| |
x x2 1 + px 3
Δ = (y − x)(z − x) 1 y+x (
p y 2 + x 2 + xy )
0 z−y p(z − y)(x + y + z)
| |
x x2 1 + px 3
= (y − x)(z − x)(z − y) 1 y+x (
p y 2 + x 2 + xy )
0 1 p(x + y + z)
Expanding along R 3, we have:
[ ( )
Δ = (x − y)(y − z)(z − x) ( − 1)(p) xy 2 + x 3 + x 2y + 1 + px 3 + p(x + y + z)(xy)
= (x − y)(y − z)(z − x) [ − pxy − px − px y + 1 + px + px y + pxy + pxyz ]
2 3 2 3 2 2
= (x − y)(y − z)(z − x)(1 + pxyz)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q13 Using properties of determinants, prove that:
| |
3a −a + b −a + c
−b + a 3b − b + c = 3(a + b + c)(ab + bc + ca)
−c + a −c + b 3c
Page 72
Answer.
| |
3a −a + b − a + c
Δ = −b + a 3b −b + c
−c + a −c + b 3c
Applying C 1 → C 1 + C 2 + C 3, we have:
| |
a+b+c −a + b −a + c
Δ= a+b+c 3b −b + c
a+b+c −c + b 3c
| |
1 −a + b −a + c
= (a + b + c) 1 3b −b + c
1 −c + b 3c
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
| |
1 −a + b −a + c
Δ = (a + b + c) 0 2b + a a−b
0 a−c 2c + a
Expanding along C 1, we have:
Δ = (a + b + c)[(2b + a)(2c + a) − (a − b)(a − c)]
[
= (a + b + c) 4bc + 2ab + 2ac + a 2 − a 2 + ac + ba − bc ]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q14 Using properties of determinants, prove that:
| |
1 1+p 1+p+q
2 3 + 2p 4 + 3p + 2q =1
3 6 + 3p 10 + 6p + 3q
Answer.
| |
1 1+p 1+p+q
Δ= 2 3 + 2p 4 + 3p + 2q
3 6 + 3p 10 + 6p + 3q
| |
3a −a + b − a + c
Δ = −b + a 3b −b + c
−c + a −c + b 3c
Applying C 1 → C 1 + C 2 + C 3, we have:
| |
a+b+c −a + b −a + c
Δ= a+b+c 3b −b + c
a+b+c −c + b 3c
| |
1 −a + b −a + c
= (a + b + c) 1 3b −b + c
1 −c + b 3c
Applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1, we have:
| |
1 −a + b −a + c
Δ = (a + b + c) 0 2b + a a−b
0 a−c 2c + a
Expanding along C 1, we have:
Δ = (a + b + c)[(2b + a)(2c + a) − (a − b)(a − c)]
[
= (a + b + c) 4bc + 2ab + 2ac + a 2 − a 2 + ac + ba − bc ]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q14 Using properties of determinants, prove that:
| |
1 1+p 1+p+q
2 3 + 2p 4 + 3p + 2q =1
3 6 + 3p 10 + 6p + 3q
Answer.
| |
1 1+p 1+p+q
Δ= 2 3 + 2p 4 + 3p + 2q
3 6 + 3p 10 + 6p + 3q
Page 73
Applying R 2 → R 2 − 2R 1 and R 3 → R 3 − 3R 1, we have:
| |
1 1+p 1+p+q
Δ= 0 1 2+p
0 3 7 + 3p
Applying R 3 → R 3 − 3R 2, , we have:
| |
1 −a + b −a + c
Δ = (a + b + c) 0 2b + a a−b
0 a−c 2c + a
Expanding along C 1, we have:
Δ = (a + b + c)[(2b + a)(2c + a) − (a − b)(a − c)]
[
= (a + b + c) 4bc + 2ab + 2ac + a 2 − a 2 + ac + ba − bc ]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q15 Using properties of determinants, prove that:
| |
sinα cosα cos(α + δ)
sinβ cosβ cos(β + δ) = 0
sinγ cosγ cos(γ + δ)
Answer.
| |
sinα cosα cos(α + δ)
Δ = sinβ cosβ cos(β + δ)
sinγ cosγ cos(γ + δ)
| |
sinαsinδ cosαcosδ cosαcosδ − sinαsinδ
1
= sin δcos δ sinβsinδ cosβcosδ cosβcosδ − sinβsinδ
sinγsinδ cosγcosδ cosγcosδ − sinγsinδ
| |
cosαcosδ cosαcosδ cosαcosδ − sinαsinδ
1
Applying C 1 → C 1 + C 3,we have : Δ = sin δcos δ cosβcosδ cosβcosδ cosβcosδ − sinβsinδ
cosγcosδ cosγcosδ cosγcosδ − sinγsinδ
Here, two columns C 1 and C 2 are identical.
∴Δ=0
Hence, the given result is proved.
| |
1 1+p 1+p+q
Δ= 0 1 2+p
0 3 7 + 3p
Applying R 3 → R 3 − 3R 2, , we have:
| |
1 −a + b −a + c
Δ = (a + b + c) 0 2b + a a−b
0 a−c 2c + a
Expanding along C 1, we have:
Δ = (a + b + c)[(2b + a)(2c + a) − (a − b)(a − c)]
[
= (a + b + c) 4bc + 2ab + 2ac + a 2 − a 2 + ac + ba − bc ]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved.
Page : 142 , Block Name : Miscellaneous Exercise
Q15 Using properties of determinants, prove that:
| |
sinα cosα cos(α + δ)
sinβ cosβ cos(β + δ) = 0
sinγ cosγ cos(γ + δ)
Answer.
| |
sinα cosα cos(α + δ)
Δ = sinβ cosβ cos(β + δ)
sinγ cosγ cos(γ + δ)
| |
sinαsinδ cosαcosδ cosαcosδ − sinαsinδ
1
= sin δcos δ sinβsinδ cosβcosδ cosβcosδ − sinβsinδ
sinγsinδ cosγcosδ cosγcosδ − sinγsinδ
| |
cosαcosδ cosαcosδ cosαcosδ − sinαsinδ
1
Applying C 1 → C 1 + C 3,we have : Δ = sin δcos δ cosβcosδ cosβcosδ cosβcosδ − sinβsinδ
cosγcosδ cosγcosδ cosγcosδ − sinγsinδ
Here, two columns C 1 and C 2 are identical.
∴Δ=0
Hence, the given result is proved.
Page 74
Page : 142 , Block Name : Miscellaneous Exercise
Q16 Solve the system of the following equations
2 3 10
x
+ y + z =4
4 6 5
x
− y + z =1
6 9 20
x
+ y − z =2
1 1 1
Answer. Let x = p, y = q, z = r
Then the given system of equations is as follows:
2p + 3q + 10r = 4
4p − 6q + 5r = 1
6p + 9q − 20r = 2
This system can be written in the form of AX = B, where
[ ] []
2 −6 5
A= 4
6
−6
9
5
− 20
,X =
p
r
and B =
[] 4
2
and B =
[]
4
2
Now ,
| A | = 2(120 − 45) − 3( − 80 − 30) + 10(36 + 36)
= 150 + 330 + 720
= 1200
Thus, A is non-singular. Therefore, its inverse exists.
A 11 = 75, A 12 = 110, A 13 = 72
Now , A 21 = 150, A 22 = − 100, A 23 = 0
A 31 = 75, A 32 = 30, A 33 = − 24
1
∴ A −1 = adjA
|A|
[ ]
75 150 75
1
= 110 − 100 30
1200
72 0 − 24
X = A − 1B
[] [ ][ ]
p 75 150 75 4
Now , 1
⇒ q = 1200 110 − 100 30 1
r 72 0 − 24 2
Q16 Solve the system of the following equations
2 3 10
x
+ y + z =4
4 6 5
x
− y + z =1
6 9 20
x
+ y − z =2
1 1 1
Answer. Let x = p, y = q, z = r
Then the given system of equations is as follows:
2p + 3q + 10r = 4
4p − 6q + 5r = 1
6p + 9q − 20r = 2
This system can be written in the form of AX = B, where
[ ] []
2 −6 5
A= 4
6
−6
9
5
− 20
,X =
p
r
and B =
[] 4
2
and B =
[]
4
2
Now ,
| A | = 2(120 − 45) − 3( − 80 − 30) + 10(36 + 36)
= 150 + 330 + 720
= 1200
Thus, A is non-singular. Therefore, its inverse exists.
A 11 = 75, A 12 = 110, A 13 = 72
Now , A 21 = 150, A 22 = − 100, A 23 = 0
A 31 = 75, A 32 = 30, A 33 = − 24
1
∴ A −1 = adjA
|A|
[ ]
75 150 75
1
= 110 − 100 30
1200
72 0 − 24
X = A − 1B
[] [ ][ ]
p 75 150 75 4
Now , 1
⇒ q = 1200 110 − 100 30 1
r 72 0 − 24 2
Page 75
[ ]
300 + 150 + 150
1
= 1200 440 − 100 + 60
288 + 0 − 48
[]
1
[]
2
600
1 1
= 1200 400 = 3
240 1
5
1 1 1
∴ p = 2 , q = 3 , and r = 5
Hence, x = 2. v = 3. and z = 5
Page : 142 , Block Name : Miscellaneous Exercise
Q17 Choose the correct answer.
If a, b, c, are in A.P., then the determinant
| |
x+2 x+3 x + 2a
x+3 x+4 x + 2b
x+4 x+5 x + 2c
A. 0 B. 1 C. x D. 2x
Answer. Answer:A
| |
x+2 x+3 x + 2a
Δ= x+3 x+4 x + 2b
x+4 x+5 x + 2c
| |
x+2 x+3 x + 2a
= x+3 x+4 x + (a + c) (2b = a + c as a, b, and c are in A.P. )
x+4 x+5 x + 2c
Applying R 1 → R 1 − R 2 and R 3 → R 3 − R 2, we have:
| |
−1 −1 a−c
Δ= x+3 x+4 x + (a + c)
1 1 c−a
300 + 150 + 150
1
= 1200 440 − 100 + 60
288 + 0 − 48
[]
1
[]
2
600
1 1
= 1200 400 = 3
240 1
5
1 1 1
∴ p = 2 , q = 3 , and r = 5
Hence, x = 2. v = 3. and z = 5
Page : 142 , Block Name : Miscellaneous Exercise
Q17 Choose the correct answer.
If a, b, c, are in A.P., then the determinant
| |
x+2 x+3 x + 2a
x+3 x+4 x + 2b
x+4 x+5 x + 2c
A. 0 B. 1 C. x D. 2x
Answer. Answer:A
| |
x+2 x+3 x + 2a
Δ= x+3 x+4 x + 2b
x+4 x+5 x + 2c
| |
x+2 x+3 x + 2a
= x+3 x+4 x + (a + c) (2b = a + c as a, b, and c are in A.P. )
x+4 x+5 x + 2c
Applying R 1 → R 1 − R 2 and R 3 → R 3 − R 2, we have:
| |
−1 −1 a−c
Δ= x+3 x+4 x + (a + c)
1 1 c−a
Page 76
Applying R 1 → R 1 + R 3, we have:
| |
0 0 0
Δ= x+3 x+4 x+a+c
1 1 c−a
Here, all the elements of the rst row (R1) are zero.
Hence, we have Δ = 0.
The correct answer is A.
Page : 143 , Block Name : Miscellaneous Exercise
Q18 Choose the correct answer.
[ ]
x 0 0
If x, y, z are nonzero real numbers, then the inverse of matrix A = 0 y 0 is
0 0 z
[ ] [ ]
x −1 0 0 x −1 0 0
A. 0 y −1 0 B. xyz 0 y −1 0
0 0 z −1 0 0 z −1
[ ] [ ]
x 0 0 1 0 0
1 1
C. xyz 0 y 0 D.
xyz
0 1 0
0 0 z 0 0 1
Answer. Answer: A
[ ]
x 0 0
A= 0 y 0
0 0 z
∴ | A | = x(yz − 0) = xyz ≠ 0
Now, A 11 = yz, A 12 = 0, A 13 = 0
A 21 = 0, A 22 = xz, A 23 = 0
A 31 = 0, A 32 = 0, A 33 = xy
[ ]
yz 0 0
∴ adjA = 0 xz 0
0 0 xy
1
∴ A −1 = adjA
|A|
| |
0 0 0
Δ= x+3 x+4 x+a+c
1 1 c−a
Here, all the elements of the rst row (R1) are zero.
Hence, we have Δ = 0.
The correct answer is A.
Page : 143 , Block Name : Miscellaneous Exercise
Q18 Choose the correct answer.
[ ]
x 0 0
If x, y, z are nonzero real numbers, then the inverse of matrix A = 0 y 0 is
0 0 z
[ ] [ ]
x −1 0 0 x −1 0 0
A. 0 y −1 0 B. xyz 0 y −1 0
0 0 z −1 0 0 z −1
[ ] [ ]
x 0 0 1 0 0
1 1
C. xyz 0 y 0 D.
xyz
0 1 0
0 0 z 0 0 1
Answer. Answer: A
[ ]
x 0 0
A= 0 y 0
0 0 z
∴ | A | = x(yz − 0) = xyz ≠ 0
Now, A 11 = yz, A 12 = 0, A 13 = 0
A 21 = 0, A 22 = xz, A 23 = 0
A 31 = 0, A 32 = 0, A 33 = xy
[ ]
yz 0 0
∴ adjA = 0 xz 0
0 0 xy
1
∴ A −1 = adjA
|A|
Page 77
[ ]
yz 0 0
1
= xyz 0 xz 0
0 0 xy
[ ]
yz
xyz 0 0
xz
= 0 xyz 0
xy
0 0 xyz
[ ]
1
x 0 0
= 0
0
1
y
0
0
1
=
[ 0
0
y −1
0 z −1
0
]
z
The correct answer is A.
Page : 143 , Block Name : Miscellaneous Exercise
Q19 Choose the correct answer.
[ ]
1 sinθ 1
Let , A = − sinθ 1 sinθ Where 0 ≤ θ ≤ 2π, , then
−1 − sinθ 1
A. Det (A) = 0
B. Det (A) ∈ (2, ∞)
C. Det (A) ∈ (2, 4)
D. Det (A) ∈ [2, 4]
Answer. Answer: D
yz 0 0
1
= xyz 0 xz 0
0 0 xy
[ ]
yz
xyz 0 0
xz
= 0 xyz 0
xy
0 0 xyz
[ ]
1
x 0 0
= 0
0
1
y
0
0
1
=
[ 0
0
y −1
0 z −1
0
]
z
The correct answer is A.
Page : 143 , Block Name : Miscellaneous Exercise
Q19 Choose the correct answer.
[ ]
1 sinθ 1
Let , A = − sinθ 1 sinθ Where 0 ≤ θ ≤ 2π, , then
−1 − sinθ 1
A. Det (A) = 0
B. Det (A) ∈ (2, ∞)
C. Det (A) ∈ (2, 4)
D. Det (A) ∈ [2, 4]
Answer. Answer: D
Page 78
[ ]
1 sinθ 1
A= − sinθ 1 sinθ
−1 − sinθ 1
( ) (
∴ | A | = 1 1 + sin 2θ − sinθ( − sinθ + sinθ) + 1 sin 2θ + 1 )
= 1 + sin 2θ + sin 2θ + 1
= 2 + 2sin 2θ
(
= 2 1 + sin 2θ ) Now
0 ≤ θ ≤ 2π
⇒ 0 ≤ sin 2θ ≤ 1
⇒ 1 ≤ 1 + sin 2θ ≤ 2
( )
⇒ 2 ≤ 2 1 + sin 2θ ≤ 4
⇒ − 1 ≤ sinθ ≤ 1
∴ Det(A) ∈ [2, 4]
Page : 143 , Block Name : Miscellaneous Exercise
1 sinθ 1
A= − sinθ 1 sinθ
−1 − sinθ 1
( ) (
∴ | A | = 1 1 + sin 2θ − sinθ( − sinθ + sinθ) + 1 sin 2θ + 1 )
= 1 + sin 2θ + sin 2θ + 1
= 2 + 2sin 2θ
(
= 2 1 + sin 2θ ) Now
0 ≤ θ ≤ 2π
⇒ 0 ≤ sin 2θ ≤ 1
⇒ 1 ≤ 1 + sin 2θ ≤ 2
( )
⇒ 2 ≤ 2 1 + sin 2θ ≤ 4
⇒ − 1 ≤ sinθ ≤ 1
∴ Det(A) ∈ [2, 4]
Page : 143 , Block Name : Miscellaneous Exercise
Document Details
| Board / Org | NCERT |
| Exam | Class 12 |
| Type | Solution |
| Pages | 78 |
| Updated | 22 Jul 2026 |