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

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

RBSE Solutions For Class 12 Maths Chapter 5 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

RBSE Solutions For Class 12 Maths Chapter 5.1 Question 2.
Solution:

RBSE Solutions For Class 12 Maths Chapter 5 Miscellaneous 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,

Cofactors of matrix A,

Cofactors of matrix A.

Ex 5.1 Class 12 Question 4.
If matrix

then find A-1 and prove that:
(i) A-1A= I3
(ii) A-1 = F(-α)
Solution:

Then, |A| = cos α (cos α – 0) + sin α (sin α – 0) + 0(0 – 0)
= cos2 α + sin2 α
|A|= 1 ≠ 0 So,
A-1 exists.
Cofactors of matrix A,

Class 12 Math Chapter 5.1 Solution Question 5.
, then prove that : A-1 = AT.
Solution:
Let

RBSE Solutions For Class 12 Maths Chapter 5.1 Question 6.
If matrix , then prove that A-1 = A3.
Solution:
Given,

So, A-1 exists.
On finding adjoint of matrix A,
a11 = – 1, a12 = – 2, a21 = 1, a22 = 1
Matrix formed by adjoint of A,

From (i) and (ii),
A-1 = A3
Hence proved.

Exercise 5.1 Class 12 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

RBSE Solutions For Class 12 Maths Chapter 5 Question 8.
If
Solution:
Given

Matrix 5.1 Question 9.
Show prove that matrix satisfies equation A2 – 6A + 17I = O. Thus find A-1.
Solution:
Given

Maths Chapter 5 Class 12 Question 10.
If matrix
, then show that A2 + 4A – 42I = O. Hence A2.
Solution:
Given

So, given matrix satisfies A2 + 4A – 42I = O
Now A2 + 4A – 42I = O
⇒ A2 + 4A = 42I
⇒ A-1 (A2 + 4A) = 42A-1.I
⇒ A-1.A2 + 4A-1.A = 42A-1.I
⇒ A + 4I = 42A-1
(∵ A-1.A = I and A-1.I = A-1)