PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.4

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 3 Matrices Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4

Direction (1 – 17) Using elementary transformations, find the inverse of each of the matrices.

Question 1.
\(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
We know that A = IA

Question 2.
\(\left[\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right]\)

We know that A = IA

Question 3.
\(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
We know that A = IA

Question 4.
\(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\)
We know that A = IA

Question 5.
\(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
Solution.
\(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
We know that A = IA

Question 6.
\(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution.
Let A = \left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]
We know that A = IA

Question 7.
\(\left[\begin{array}{ll}
\mathbf{3} & 1 \\
5 & 2
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
\mathbf{3} & 1 \\
5 & 2
\end{array}\right]\)

We know that A = IA

Question 8.
\(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
We know that A = IA

Question 9.
\(\left[\begin{array}{cc}
3 & 10 \\
2 & 7
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{cc}
3 & 10 \\
2 & 7
\end{array}\right]\)
We know that A = IA

Question 10.
\(\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]\)
Solution.
\(\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]\)
We know that A = IA.

Question 11.
\(\left[\begin{array}{rr}
2 & -6 \\
1 & -2
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{rr}
2 & -6 \\
1 & -2
\end{array}\right]\)
We know that A = IA.

Question 12.
\(\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]\)
We know that A = IA.

Now, in the above equation, we can see all the zeroes in the second law of the matrix on the L.H.S.
Threrefore, A^-1 does not exist.

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

We know that A = IA.

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right] A\) (applying R₁ → R₁ + R₂)

∴ A^-1 = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\)

Question 14.
\(\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]\)
We know that A = IA
∴ \(\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] A\)

Applying R₁ → R₁ – \(\frac{1}{2}\) R₂, we have
\(\left[\begin{array}{ll}
0 & 0 \\
4 & 2
\end{array}\right]=\left[\begin{array}{rr}
1 & -\frac{1}{2} \\
0 & 1
\end{array}\right] A\)

Now, in the above equation, we can see all the zeroes in the first row of the matrix on the L.H.S.
Therefore, A^-1 does not exist.

Question 15.
\(\left[\begin{array}{rrr}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
Solution.
let A = \(\left[\begin{array}{rrr}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
We know that A = IA

Question 16.
\(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
We know that A = IA

Question 17.
\(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
We know that A = IA

Question 18.
Matrices A and B will be inverse of each other only if
(A) AB = BA
(B) AB = BA = 0
(C) AB = 0,BA = I
(D) AB = BA = I
Solution.
We know that if A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is said to be the inverse of A.
In this case, it is clear that A is the inverse of B.
Thus, matrices A and B will be inverse of each other only if AB = BA = I. Hence, the correct answer is (D).