# Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 3

### Question 25. Show that the matrix A = satisfies the equation A^{3 }– A^{2 }– 3A – I_{3 }= 0. Hence, find A^{-1}.

**Solution:**

Here, A =

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^{2}=A

^{3}=Now A

^{3 }– A^{2 }– 3A – I_{3 }==

=

So, A

^{3 }– A^{2 }– 3A – I_{3 }= 0⇒ A

^{-1}(AAA) – A^{-1}(AA) – 3A^{-1}A – A^{-1}I = 0⇒ A

^{2 }– A – 3I – A^{-1 }= 0⇒ A

^{-1 }= A^{2 }– A – 3I=

Therefore, A

^{-1}=

### Question 26. If A = . Verify that A^{3 }– 6A^{2 }+ 9A – 4I = O and hence find A^{-1}.

**Solution:**

Here, A =

A

^{2}==

=

A

^{3}= A^{2}A ==

=

Now, A

^{3 }– 6A^{2 }+ 9A – 4I=

=

=

=

So, A

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

^{-1}⇒ A

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

^{-1}⇒ 4A

^{-1 }= A^{2}I – 6AI + 9I=

=

=

⇒ A

^{-1 }=

### Question 27. If A =, prove that A^{-1 }= A^{T}.

**Solution:**

Here, A =

A

^{T}=Now, Finding A

^{-1}|A| = 1/9[-8(16 + 56) – 1(16 – 7) + 4(-32 – 4)]

= -81

Therefore, inverse of A exists

Cofactors of A are:

C

_{11 }= 72 C_{12}= -9 C_{13 }= -36C

_{21 }= -36 C_{22}= -36 C_{23 }= -63C

_{31 }= -9 C_{32 }= 72 C_{33 }= -36adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, A

^{-1}==

= A

^{T}Hence Proved

**Question 28. If A = ****, show that A**^{-1 }= A^{3}.

^{-1 }= A

^{3}.

**Solution:**

Here, A =

|A| = 3(-3 + 4) + 3(2 – 0) + 4(-2 – 0)

= 3 + 6 – 8

= 1

Therefore, inverse of A exists

Cofactors of A are:

C

_{11 }= 1 C_{12}= -2 C_{13 }= -2C

_{21 }= -1 C_{22}= 3 C_{23 }= 3C

_{31 }= 0 C_{32 }= -4 C_{33 }= -3adj A =

A

^{-1}= 1/|A|. adj A=

Now, A

^{2}==

A

^{3}==

= A

^{-1}Hence Proved

### Question 29. If A =, show that A^{2 }= A^{-1}.

**Solution: **

Here, A =

LHS = A

^{2}=

=

|A| = -1(1 – 0) – 2(-1 – 0) + 0

= 1

Therefore, inverse of A exists

Cofactors of A are:

C

_{11 }= -1 C_{12}= 0 C_{13 }= -1C

_{21 }= 0 C_{22}= 0 C_{23 }= 1C

_{31 }= 21 C_{32 }= 1 C_{33 }= 1adj A =

=

A

^{-1 }= 1/|A|. adj AHence, A

^{-1}==

= A

^{2 }Hence Proved

### Question 30. Solve the matrix equation, where X is a 2×2 matrix.

**Solution:**

We have,

Let A = and B =

So. AX = B

⇒ X = A

^{-1}BNow, |A| = 5 – 4 = 1

Co factors of A are:

C

_{11}= 1 C_{12}= -1C

_{21}= -4 C_{22}= 5adj A =

=

A

^{-1}=Therefore, X =

X =

### Question 31. Find the matrix X satisfying the matrix equation: X.

**Solution:**

We have,

Let A =and B =

So, XA = B

XAA

^{-1}= BA^{-1}XI = BA

^{-1 }………..(i)^{ }Now, |A| = -7

Co factors of A are:

C

_{11}= -2 C_{12}= 1C

_{21}= -3 C_{22}= 5adj A =

=

A

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

Therefore, X =

=

X =

### Question 32. Find the matrix X for which: X

**Solution:**

Let, A =

B =

C =

Then the given equation becomes

A × B = C

⇒ X = A

^{-1}CB^{-1}Now |A| = 35 -14 = 21

|B| = -1 + 2 = 1

A

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

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

^{-1}CB^{-1}==

### Question 33. Find the matrix X satisfying the equation:

**Solution:**

Let A = B =

AXB = I

X = A

^{-1}B^{-1}|A| = 6 – 5 = 1

|B| = 10 – 9 = 1

A

^{-1}= adj A /|A| =B

^{-1}= adj B/|B| =X =

=

### Question 34. If A = , Find A^{-1} and prove that A^{2 }– 4A – 5I = O.

**Solution:**

Here, A =

A

^{2}==

Now, A

^{2 }+ 4A – 5I=

=

Now, A

^{2 }– 4A – 5I = O⇒ A

^{-1}AA – 4A^{-1}A – 5A^{-1}I = O⇒ 5A

^{-1}= [A – 4I]=

=

A

^{-1 }=

### Question 35. If A is a square matrix of order n, prove that |A adj A| = |A|^{n}.

**Solution:**

Given, |A adj A| = |A|

^{n}Taking LHS = |A Adj A|

= |A| |Adj A|

= |A| |A|

^{n-1}= |A|

^{n-1+1}= |A|

^{n}= RHSHence Proved

### Question 36. If A^{-1} = and B = , find (AB)^{-1}.

**Solution:**

Here, B =

|B| = 1(3 – 0) – 2(-1 – 0) – 2(2 – 0)

= 3 + 2 – 4 = 1

Therefore, inverse of B exists

Cofactors of B are:

C

_{11 }= 3 C_{12}= 1 C_{13 }= 2C

_{21 }= 2 C_{22}= 1 C_{23 }= 2C

_{31 }= 6 C_{32 }= 2 C_{33 }= 5adj A =

=

Therefore,

B

^{-1 }=Hence, (AB)

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

=

### Question 37. If A = , find (A^{T})^{-1}.

**Solution:**

Assuming B = A

^{T}=|B| = 1(-1 – 8) – 0 – 2(-8 + 3)

= -9 + 10 = 1

Therefore, inverse of B exists

Cofactors of B are:

C

_{11 }= -9 C_{12}= 8 C_{13 }= -5C

_{21 }= -8 C_{22}= 7 C_{23 }= -4C

_{31 }= -2 C_{32 }= 2 C_{33 }= -1adj B =

=

B

^{-1}=or (A

^{T})^{-1}=

### Question 38. Find the adjoint of the matrix A = and hence show that A (adj A) = |A|I_{3}.

**Solution:**

Here, A =

|A| = -1(1 – 4) – 2(2 + 4) – 2(-4 – 2)

= 3 + 12 + 12 = 27

Therefore, inverse of A exists

Cofactors of A are:

C

_{11 }= -3 C_{12}= -6 C_{13 }= -6C

_{21 }= 6 C_{22}= 3 C_{23 }= -6C

_{31 }= 6 C_{32 }= -6 C_{33 }= 3adj A =

=

A (adj A) =

=

or A (adj A) = 27

Hence, A (adj A) = |A|I

_{3 }Hence Proved

### Question 39. If A = , A^{-1} and show that A^{-1} = 1/2(A^{2} – 3I).

**Solution:**

Here, A =

|A| = 0 – 1(0 – 1) + 1(1 – 0)

= 1 + 1 = 2

Therefore, inverse of A exists

Cofactors of A are:

C

_{11 }= -1 C_{12}= 1 C_{13 }= 1C

_{21 }= 1 C_{22}= -1 C_{23 }= 1C

_{31 }= 1 C_{32 }= 1 C_{33 }= -1adj A =

=

A

^{-1 }= 1/|A|. adj AHence, A

^{-1}=Now, A

^{2}– 3I ==

=

Hence, A

^{-1}= 1/2(A^{2}– 3I)Hence Proved