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: