# Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Exercise 4.5

### Find adjoint of each of the matrices in Exercises 1 and 2.

### Question 1.

**Solution: **

A =

A

_{11}= 4; A_{12}= -3; A_{21}= -2; A_{22}= 1adj A =

adj A =

### Question 2.

**Solution: **

A =

A

_{11}=A

_{11}= 3 – 0 = 3A

_{12}=A

_{12}= -(2 + 10) = -12A

_{13}=A

_{13}= 0 + 6 = 6A

_{21}=A

_{21}= -(-1 – 0) = 1A

_{22}=A

_{22}= 1 + 4 = 5A

_{23}=A

_{23}= -(0 – 2) = 2A

_{31}=A

_{31}= -5 – 6 = -11A

_{32}=A

_{32}= -(5 – 4) = -1A

_{33}=A

_{33}= 3 + 2 = 5adj A =

adj A =

### Verify A(adj A) = (adj A)A = |A| I in exercises 3 and 4.

### Question 3.

**Solution: **

|A| = -12 -(-12)

|A| = -12 + 12 = 0

so, |A|*I = 0 *

|A|*I = ……… (1)

Now, for adjoint of A

A

_{11}= -6A

_{12}= 4A

_{21}= -3A

_{22}= 2adj A =

adj A =

Now, A(adj A) =

A(adj A) = [Tex]\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]

A(adj A) = …………(2)

Now, (adj A)A =

(adj A)A =

(adj A)A = …………….(3)

From eq(1), (2), and (3), you can see that A(adj A) = (adj A)A = |A|I

### Question 4.

**Solution:**

A =

|A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0)

|A| = 11

|A| * I =

|A| * I =

Now, for adjoint of A

A

_{11}= 0A

_{12}= -(9 + 2) = -11A

_{13}= 0A

_{21}= -(-3 – 0) = 3A

_{22}= 3 – 2 = 1A

_{23}= -(0 + 1) = -1A

_{31}= 2 – 0 = 2A

_{32}= -(-2 – 6) = 8A

_{33}= 0 + 3 = 3adj A =

Now,

A(adjA) =

A(adj A) =

A(adj A) =

Also,

(adj A).A =

(adj A).A =

(adj A).A =

From above, you can see,

A(adj A) = (adj A)A = I

### Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

### Question 5.

**Solution:**

|A| = 6 – (-8) = 14

|A| â‰ 0, So inverse exists.

A

_{11}= 3A

_{12}= 2A

_{21}= -4A

_{22}= 2adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=

### Question 6.

**Solution: **

A =

|A| = -2 + 15 = 13 â‰ 0

Hence, inverse exists.

A

_{11}= 2A

_{12}= 3A

_{21}= -5A

_{22}= -1adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=

### Question 7.

**Solution:**

A =

|A| = 1(10 – 0) – 2(0 – 0) + 3(0 – 0) = 10

For adj A

A

_{11 }= 10 – 0 = 0A

_{12 }= -(0 – 0) = 0A

_{13 }= 0 – 0 = 0A

_{21 }= -(10 – 0) = -10A

_{22 }= 5 – 0 = 5A

_{23 }= -(0 – 0) = 0A

_{31 }= 8 – 6 = 2A

_{32 }= -(4 – 0) = -4A

_{33 }= 2 – 0 = 2adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=

### Question 8.

**Solution:**

A =

|A| = 1(-3 – 0) – 0 + 0 = -3 â‰ 0

Hence, inverse exists.

For adj A

A

_{11}= -3 – 0 = -3A

_{12}= -(-3 – 0) = 3A

_{13}= 6 – 15 = -9A

_{21}= -(0 – 0) = 0A

_{22}= -1 – 0 = -1A

_{23}= -(2 – 0) = -2A

_{31}= 0 – 0 = 0A

_{32}= -(0 – 0) = 0A

_{33}= 3 – 0 = 3adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=

### Question 9.

**Solution:**

A =

|A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -3 â‰ 0

Hence, inverse exists.

For adj A

A

_{11}= -1 – 0 = -1A

_{12}= -(4 – 0) = -4A

_{13}= 8 – 7 = 1A

_{21}= -(1 – 6) = 5A

_{22}= 2 + 21 = 23A

_{23}= -(4 + 7) = -11A

_{31}= 0 + 3 = 3A

_{32}= -(0 – 12) = 12A

_{33}= -2 – 4 = -6adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=

### Question 10.

**Solution:**

A =

|A| = 1(8 – 6) – 0 + 3(3 – 4) = -1

Now for adj A

A

_{11}= 8 – 6 =2A

_{12}= -(0 + 9) = -9A

_{13}= 0 – 6 = -6A

_{21}= -(-4 + 4) =0A

_{22}= 4 – 6 = -2A

_{23}= -(-2 + 3) = -1A

_{31}= 3 – 4 = -1A

_{32}= -(-3 – 0) = 3A

_{33}= 2 – 0 = 2adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=A

^{-1}=

### Question 11.

**Solution: **

A =

|A| = 1(-cos2

Î±– sin2Î±) = -1Now,

A

_{11}= -cos2Î±– sin2Î±= -1A

_{12}= 0A

_{13}= 0A

_{21}= 0A

_{22}= -cosÎ±A

_{23}= -sinÎ±A

_{31}= 0A

_{32}= -sinÎ±A

_{33}= cosÎ±adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=A

^{-1}=

### Question 12. Let A = and B = , verify that (AB) ^{– 1 }= B ^{– 1}A ^{– 1 }

**Solution:**

A =

|A| = 15 – 14 = 1

A

_{11}= 5A

_{12}= -2A

_{21}= -7A

_{22}= 3A

^{-1}= (adj A)/|A|A

^{-1}=B =

|B| = 54 – 56 = -2

adj B =

B

^{-1}= (adj B)/|B|B

^{-1}=B

^{-1}=Now,

B

^{-1}A^{-1}=B

^{-1}A^{-1}=B

^{-1}A^{-1}=Now, AB =

AB =

AB =

|AB| = 67 * 61 – 87 * 47 = -2

adj (AB) =

(AB)

^{-1}= (adj AB)/|AB|(AB)

^{-1}=(AB)

^{-1}=From above, you can see that (AB)

^{-1 }= B^{-1}A^{-1}.Hence, it is proved.

### Question 13. A = , show that A^{2 }– 5A + 7I = O. Hence find A^{-1}.

**Solution: **

A =

A

^{2}=A

^{2}=A

^{2}=So, A

^{2}– 5A + 7I= – 5+ 7

= – +

=

= O

Hence, A

^{2}– 5A + 7I = OIt can be written as

A.A – 5A = -7I

Multiplying by A

^{-1}in both sidesA.A(A

^{-1}) – 5AA^{-1}= 7IA^{-1}A(AA

^{-1}) – 5I = -7A^{-1}AI – 5I = -7A

^{-1}A

^{-1}= -(A – 5I)/7A

^{-1}=1/7( – )A

^{-1}=

### Question 14. For the matrix A =, find the numbers a and b such that A2 + aA + bI = O.

**Solution: **

A =

A

^{2}=A

^{2}=A2 =

Now,

A

^{2}– aA + bI = OMultiplying by A

^{-1}in both sides(AA)A

^{-1}+ aAA^{-1}+ bIA^{-1}= OA(AA

^{-1}) + aI + b(IA^{-1}) = OAI + aI + bA

^{-1}= OA + aI = -bA

^{-1}A

^{-1}= -(A + aI)/bNow,

A

^{-1}= (adj A)/|A|A

^{-1}=Now,

= -1/b

=

On comparing elements you will get

-1/b = -1

b = 1

(-3 – a)/b = 1

-3 – a = 1

a = -4

Hence, a = -4 and b = 1

### Question 15. A = , show that A^{3} – 6A^{2} + 5A + 11I = O. Hence find A^{-1}

**Solution:**

A =

A

^{2}=A

^{2}=A

^{2}=A

^{3}= A^{2}.AA

^{3}=A

^{3}=A

^{3}=A

^{3}– 6A^{2}+ 5A + 11I– 6+ 5 + 11

= – + +

=

= O

Hence, A

^{3}– 6A^{2}+ 5A + 11I = ONow,

A

^{3}– 6A^{2}+ 5A + 11I = O(AAA)A

^{-1}– 6(AA)A^{-1}+ 5(AA^{-1}) + 11IA^{-1}= OAA(AA

^{-1}) – 6A(AA^{-1}) + 5(AA^{-1}) = -11(IA^{-1})A

^{2}– 6A + 5I = -11 A^{-1}A

^{-1}= -1/11(A2 – 6A + 5I) ………….(1)Now, A

^{2 }– 6A + 5I= – 6 + 5

= – +

=

From eq(1) you have

A

^{-1}= -1/11

### Question 16. A = , verify that A^{3} – 6A^{2} + 9A – 4I = O and hence fin A^{-1}.

**Solution:**

A =

A

^{2}=A

^{2}=A

^{2}=A

^{3}= A^{2}.AA

^{3}=A

^{3}=A

^{3}=Now,

A

^{3 }– 6A^{2 }+ 9A – 4I– 6 + 9– 4

= – + –

= –

=

= O

So, A

^{3}– 6A^{2}+ 9A – 4I = ONow,

A

^{3}– 6A^{2}+ 9A – 4I = OMultiplying by A

^{-1}in both sides(AAA)A

^{-1}– 6(AA)A^{-1}+ 9AA^{-1}– 4IA^{-1}= OAA(AA

^{-1}) – 6A(AA^{-1}) + 9(AA^{-1}) = 4(IA^{-1})AAI – 6AI +9I = 4A

^{-1}A

^{2}– 6A + 9I = 4A^{-1}A

^{-1}= 1/4(A2 – 6A + 9I) ……….(1)A

^{2}– 6A + 9I= – 6 + 9

=

From eq(1), you have

A

^{-1}=

### Question 17. Let A be a non-singular matrix of order 3 * 3. Then |adj A| is equal to

### (A) |A| (B) |A|^{2 }(C) |A|^{3 }(D) 3|A|

**Solution: **

You know,

(adj A)A = |A|I

(adj A)A =

|(adj A)A| =

|(adj A)A| = |A|

^{3}|(adj A)A| = |A|

^{3 }I|adj A| = |A|

^{3}Hence, option B is correct.

### Question 18. If A is an invertible matrix of order 2, then det(A^{-1}) is equal to

### (A) det(A) (B) 1/(det A) (C) 1 (D) 0

**Solution:**

Since A is an invertible matrix then A

^{-1 }exists.And A

^{-1}= (adj A)/|A|Suppose a 2 order matrix is A =

Then |A| = ad – bc

and adj A =

A

^{-1}= (adj A)/|A|A

^{-1}=A

^{-1}=|A

^{-1}| =|A

^{-1}| =|A

^{-1}| = (ad – bc)/|A|^{2}|A

^{-1}| = |A|/|A|^{2}|A

^{-1}| = 1/|A|det(A

^{-1}) = 1/(det A)Hence, option B is correct.

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