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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:$\begin{bmatrix} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{bmatrix}$
✓

Answer

$\text{A}=\begin{bmatrix} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{bmatrix}$We have A = IA
$\Rightarrow\begin{bmatrix} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & -3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -2 & 0 & 0 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-3\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-2\text{R}_1\Big]$
$\Rightarrow\begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & -3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ \frac{3}{2} & -\frac{1}{2} & 0 \\ -2 & 0 & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\Rightarrow\frac{-1}{2}\text{R}_2\Big]$
$\Rightarrow\begin{bmatrix} 1 & 1 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & -\frac{11}{1} \end{bmatrix}=\begin{bmatrix} -\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} & 0 \\ -\frac{7}{2} & \frac{1}{2} & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1-\text{R}_2\text{ and R}_3\rightarrow\text{R}_3-\text{R}_2\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} & 0 \\ -\frac{7}{11} & \frac{-1}{11} & \frac{-2}{11} \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_3\rightarrow-\frac{2}{11}\text{ R}_3\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -\frac{-2}{11} & \frac{5}{11} & \frac{-1}{11} \\ \frac{-1}{11} & -\frac{-3}{11} & \frac{5}{11} \\ \frac{7}{11} & \frac{-1}{11} & \frac{-2}{11} \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\rightarrow\text{R}_2-\frac{5}{2}\text{R}_3\text{ and R}_1\rightarrow\text{R}_1+\frac{1}{2}\text{R}_3\Big]$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{11}\begin{bmatrix} -2 & 5 & -1 \\ -1 & -3 & 5 \\ 7 & -1 & -2 \end{bmatrix}$

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