#### Need solution for RD Sharma maths class 12 chapter Adjoint and inverse of matrix exercise 6.2 question 11

Answer: $\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right]$

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

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

Solution:

Let,

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

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

Applying

\begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-3 R_{1} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & \frac{5}{2} & \frac{5}{2} \\ 0 & \frac{5}{2} & \frac{-7}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-1}{2} & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{array}\right] A \end{aligned}

Applying  $R_{2} \rightarrow \frac{2}{5} R_{2}$

$\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 1 \\ 0 & \frac{5}{2} & \frac{-7}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ \frac{-3}{2} & 0 & 1 \end{array}\right] A$

Applying

\begin{aligned} &R_{1} \rightarrow R_{1}+\frac{1}{2} R_{2} \\ &R_{3} \rightarrow R_{3}-\frac{5}{2} R_{2} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & -6 \end{array}\right]=\left[\begin{array}{ccc} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ -1 & -1 & 1 \end{array}\right] A \end{aligned}

Applying $R_{3} \rightarrow \frac{-1}{6} R_{3}$

$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right] A$

Applying

\begin{aligned} &R_{2} \rightarrow R_{2}-R_{3} \\ &R_{1} \rightarrow R_{1}-2 R_{3} \\ &\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right] A \end{aligned}

So,$A^{-1}=\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right]$