## Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.1

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Question 1.

For which value of x, matrix

is singular ?

Solution:

⇒ 1 (- 6 – 2) + 2 (-3 – x) + 3(2 – 2x) = 0

⇒ -8 – 6 – 2x + 6 – 6x = 0

⇒ -8 = 8

⇒ x = \(\frac { 8 }{ -8 }\) = -1

Hence, x = -1

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Question 2.

If matrix , then find adj.A, Also prove that, A(adj.A) = |A| I_{3} = (adj.A)A.

Solution:

Matrix made from adjoint of matrix A,

Question 3.

Find the inverse matrix of the following matrix:

Solution:

(i) Let

= 1(1 + 3) – 2(-1 + 2) + 5(3 + 2)

= 4 – 2 + 25

|A| = 27 ≠ 0

So. A^{-1} exists.

Cofactors of matrix A,

Matrix made from adjoint of matrix A,

Cofactors of matrix A,

Matrix made from adjoint of matrix A,

Cofactors of matrix A.

Matrix made from adjoint of matrix A,

Question 4.

If matrix

then find A^{-1} and prove that:

(i) A^{-1}A= I3

(ii) A^{-1} = F(-α)

(iii) A(adj.A) = |A|I = (adj A).A

Solution:

Then, |A| = cos α (cos α – 0) + sin α (sin α – 0) + 0(0 – 0)

= cos^{2} α + sin^{2} α

|A|= 1 ≠ 0 So,

A^{-1} exists.

Cofactors of matrix A,

Matrix made from adjoint of matrix A

So, A(adj.A) = |A|I = (adj.A) A Hence Proved.

Question 5.

, then prove that : A^{-1} = A^{T}.

Solution:

Let

Question 6.

If matrix , then prove that A^{-1} = A^{3}.

Solution:

Given,

So, A^{-1} exists.

On finding adjoint of matrix A,

a_{11} = – 1, a_{12} = – 2, a_{21} = 1, a_{22} = 1

Matrix formed by adjoint of A,

From (i) and (ii),

A^{-1} = A^{3}

Hence proved.

Question 7.

then find (AB)^{-1}.

Solution:

Given

Then, |A| = 5(3 – 4) – 0(2 – 2) + 4(4 – 3)

= – 5 – 0 + 4

|A| = -1 ≠ 0

So, A^{-1} exists.

On finding adjoint of matrix A,

Matrix formed by adjoint of A

Question 8.

If

Solution:

Given

Question 9.

Show prove that matrix satisfies equation A^{2} – 6A + 17I = O. Thus find A^{-1}.

Solution:

Given

Question 10.

If matrix

, then show that A^{2} + 4A – 42I = O. Hence A^{2}.

Solution:

Given

So, given matrix satisfies A^{2} + 4A – 42I = O

Now A^{2} + 4A – 42I = O

⇒ A^{2} + 4A = 42I

⇒ A^{-1} (A^{2} + 4A) = 42A^{-1}.I

⇒ A^{-1}.A^{2} + 4A^{-1}.A = 42A^{-1}.I

⇒ A + 4I = 42A^{-1}

(∵ A^{-1}.A = I and A^{-1}.I = A^{-1})

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