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

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
Question
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} 2 & -1 & 4 \\ 4 & 0 & 7 \\ 3 & -2 & 7 \end{bmatrix}$
✓

Answer

$\text{A}=\begin{bmatrix} 2 & -1 & 4 \\ 4 & 0 & 7 \\ 3 & -2 & 7 \end{bmatrix}$We know A = IA
$\Rightarrow\begin{bmatrix} 2 & -1 & 4 \\ 4 & 0 & 7 \\ 3 & -2 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & -\frac{1}{2} & 2 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1\rightarrow\frac{1}{2}\text{R}_1\Big]$
$\begin{bmatrix} 1 & -\frac{1}{2} & 2 \\ 0 & 2 & -6 \\ 0 & -\frac{1}{2} & 0 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ -2 & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\rightarrow\text{R}_2-4\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-3\text{R}_1\Big]$
$\Rightarrow\begin{bmatrix} 1 & -\frac{1}{2} & 2 \\ 0 & 2 & -3 \\ 0 & -\frac{1}{2} & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ -1 & \frac{1}{2} & 0 \\ \frac{-3}{2} & 0 & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\Rightarrow\frac{1}{2}\text{R}_2\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & -3 \\ 0 & 0 & -\frac{1}{2} \end{bmatrix}=\begin{bmatrix} 0 & \frac{1}{4} & 0 \\ -1 & \frac{1}{2} & 0 \\ -2 & \frac{1}{4} & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1\rightarrow\text{R}_1+\frac{1}{2}\text{R}_2\text{ and R}_3\rightarrow\text{R}_3+\frac{1}{2}\text{R}_2\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & \frac{1}{4} & 0 \\ -1 & \frac{1}{2} & 0 \\ -4 & \frac{1}{2} & -2 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_3\rightarrow-2\text{R}_3\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ -4 & -\frac{1}{2} & -2 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1\rightarrow\text{R}_1-\frac{1}{2}\text{R}_3\text{ and R}_2\rightarrow\text{R}_2+3\text{R}_3\Big]$
$\Rightarrow\ \text{A}^{-1}=\begin{bmatrix} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ -4 & -\frac{1}{2} & -2 \end{bmatrix}$

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