GSEB Solutions Class 12 Maths Chapter 3 Matrices Ex 3.4

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 3 Matrices Ex 3.4 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 3 Matrices Ex 3.4

Using elementary transformations, find the inverse each of following matrices, if it exists in questions 1 to 17:

Question 1.
\(\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]\)
Solution:

Question 2.
\(\left[\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}\right]\)
Solution:

Question 3.
\(\left[\begin{array}{cc} 1 & 3 \\ 2 & 7 \end{array}\right]\)
Solution:

Question 4.
\(\left[\begin{array}{cc} 2 & 3 \\ 5 & 7 \end{array}\right]\)
Solution:

Question 5.
\(\left[\begin{array}{cc} 2 & 1 \\ 7 & 4 \end{array}\right]\)
Solution:

Question 6.
\(\left[\begin{array}{cc} 2 & 5 \\ 1 & 3 \end{array}\right]\)
Solution:

Question 7.
\(\left[\begin{array}{cc} 3 & 1 \\ 5 & 2 \end{array}\right]\)
Solution:

Question 8.
\(\left[\begin{array}{cc} 4 & 5 \\ 3 & 4 \end{array}\right]\)
Solution:

Question 9.
\(\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]\)
Solution:

Question 10.
\(\left[\begin{array}{cc} 3 & -1 \\ -4 & 2 \end{array}\right]\)
Solution:

Question 11.
\(\left[\begin{array}{cc} 2 & -6 \\ 1 & -2 \end{array}\right]\)
Solution:

Question 12.
\(\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\)
Solution:
A = I₂A ⇒ \(\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\) = \(\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\)A.
Operating R₁ ↔ R₂ ⇒ \(\left[\begin{array}{cc} -2 & 1 \\ 6 & -3 \end{array}\right]\) = \(\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\)A
Operating R₂ → R₂ + 2R₁ ⇒ \(\left[\begin{array}{cc} -2 & 1 \\ 0 & 0 \end{array}\right]\) = \(\left[\begin{array}{cc} 0 & 1 \\ 1 & 3 \end{array}\right]\)A

Question 13.
\(\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\)
Solution:

Question 14.
\(\left[\begin{array}{cc} 2 & 1 \\ 4 & 2 \end{array}\right]\)
Solution:
A = I₂A ⇒ \(\left[\begin{array}{cc} 2 & 1 \\ 4 & 2 \end{array}\right]\) = \(\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\)A.
Operating R₂ → R₂ – 2R₁ ⇒ \(\left[\begin{array}{cc} 2 & 1 \\ 0 & 0 \end{array}\right]\) = \(\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right]\)A
In the second row of L.H.S., each element is zero.
∴ A^-1 does nor exist.

Question 15.
\(\left[\begin{array}{ccc} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]\)
Solution:

Question 16.
\(\left[\begin{array}{ccc} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]\)
Solution:

Question 17.
\(\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]\)
Solution:

Question 18.
Matrices A and B will be inverse of each other only, if
(A) AB = BA
(B) AB = BA = O
(C) AB = O, BA = I
(D) AB = BA = I
Solution:
If B is the inverse of A, then AB = BA = t.
∴ Part (D) is the correct answer.