#### Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 9

Answer: $\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right]$

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: $\left[\begin{array}{rrr} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]$

Solution: Let $A=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]$

\begin{aligned} &A=I A \\ &A=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}

Applying $R_{1} \rightarrow \frac{1}{3} R_{1}$

$\Rightarrow\left[\begin{array}{lll} 1 & -1 & \frac{4}{3} \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$

Applying $R_{2} \rightarrow R_{2}-2 R_{1}$

$\Rightarrow\left[\begin{array}{ccc} 1 & -1 & \frac{4}{3} \\ 0 & -1 & \frac{4}{3} \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$

Applying $R_{2} \rightarrow-R_{2}$

$\Rightarrow\left[\begin{array}{ccc} 1 & -1 & \frac{4}{3} \\ 0 & 1 & \frac{-4}{3} \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{2}{3} & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$

Applying

\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2} \\ &R_{3} \rightarrow R_{3}+R_{2} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & \frac{-4}{3} \\ 0 & 0 & \frac{-1}{3} \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ \frac{2}{3} & -1 & 0 \\ \frac{2}{3} & -1 & 1 \end{array}\right] A \end{aligned}

Applying $R_{3} \rightarrow-3 R_{3}$

$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] A$

Applying $R_{2} \rightarrow R_{2}+\frac{4}{3} R_{3}$

$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] A$

So, $A^{-1}=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right]$