Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.1
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 = = -1
Hence, x = -1
Question 2.
If matrix , then find adj.A, Also prove that, A(adj.A) = |A| I3 = (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-1A= 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)
= cos2 α + sin2 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 = AT.
Solution:
Let
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.
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 A2 – 6A + 17I = O. Thus find A-1.
Solution:
Given
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)
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