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Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
Question
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$
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Answer

Let $\text{A}=\begin{bmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$ To find inverse, first write A = IA. i.e., $\begin{bmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$$\Rightarrow\begin{bmatrix} 1 & -1 & -2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_1\rightarrow(-1)\text{R}_1\big]$
$\Rightarrow\begin{bmatrix} 1 & -1 & -2 \\ 0 & 3 & 5 \\ 0 & 4 & 7 \end{bmatrix}=\begin{bmatrix} -1 & 0 & 0 \\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-3\text{R}_1\big]$
$\begin{bmatrix} 1 & -1 & -2 \\ 0 & 1 & \frac{5}{3} \\ 0 & 4 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 3 & 0 & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\rightarrow\frac{1}{3}\text{R}_2\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix} \frac{-2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{5}{3} & -\frac{4}{3} & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_3\rightarrow\text{R}_3-4\text{R}_2\text{ and R}_1\rightarrow\text{R}_1+\text{R}_2\big]$
$\begin{bmatrix} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{-2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 5 & -4 & 3 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_3\rightarrow3\text{R}_3\big]$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\rightarrow\text{R}_2-\frac{5}{3}\text{R}_3\text{ and R}_1\rightarrow\text{R}_1+\frac{1}{3}\text{R}_3\Big]$
Hence, $\text{A}^{-1}=\begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{bmatrix}$

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