OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Ex 6(i)

Students often turn to ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(i) to clarify doubts and improve problem-solving skills.

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(i)

Using elementary transformation, find the inverse of the following matrices, if it exists.

Question 1.
A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
Since A = IA ⇒ \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)A
Operate R₂ → R₂ – 2R₁
\(\left[\begin{array}{ll}
1 & -1 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right]\)A
R₂ → \(\frac { 1 }{ 5 }\)R₂
\(\left[\begin{array}{cc}
1 & -1 \\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 / 5 & 1 / 5
\end{array}\right]\)A
Operate R₂ → R₁ + R₂
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
3 / 5 & 1 / 5 \\
-2 / 5 & 1 / 5
\end{array}\right]\)A
∴ A^-1 = \(\left[\begin{array}{cc}
3 / 5 & 1 / 5 \\
-2 / 5 & 1 / 5
\end{array}\right]\)
[using def. of inverse, A^-1A = I]

Question 2.
A = \(\left[\begin{array}{rr}
1 & 2 \\
2 & -1
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{rr}
1 & 2 \\
2 & -1
\end{array}\right]\)
Since A = IA ⇒ \(\left[\begin{array}{rr}
1 & 2 \\
2 & -1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)A
Operate R₂ → R₂ – 2R₁
\(\left[\begin{array}{cc}
1 & 2 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right]\)A
Operate R₂ → \(\frac { -1 }{ 5 }\)R₂
\(\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
2 / 5 & -1 / 5
\end{array}\right]\)A
Operate R₁ → R₁ – 2R₂
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 / 5 & 2 / 5 \\
2 / 5 & -1 / 5
\end{array}\right]\)A
∴ A^-1 = \(\left[\begin{array}{cc}
1 & 2 / 5 \\
2 / 5 & -1 / 5
\end{array}\right]\)
[using def. of inverse, A^-1A = I]

Question 3.
\(\left[\begin{array}{ll}
9 & 5 \\
7 & 4
\end{array}\right]\)
Solution:

Question 4.
\(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]\)
Since A = IA ⇒ \(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)A
Operate R₁ → R₁ – 2R₂
\(\left[\begin{array}{cc}
0 & 0 \\
-5 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]\)A
Since these are all zeros in R₁ of left side matrix
Thus A^-1 does not exists.

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

Question 6.
\(\left[\begin{array}{rrr}
1 & -1 & 0 \\
2 & 5 & 3 \\
0 & 2 & 1
\end{array}\right]\)
Solution:

Question 7.
\(\left[\begin{array}{rrr}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution: