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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:$\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$
✓

Answer

$\text{A}=\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$We know A = IA
$\Rightarrow\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1\rightarrow\frac{1}{2}\text{R}_1\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-5\text{R}_1\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2}\\ 0 & 0 &\frac{1}{2} \end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1\end{bmatrix}\text{A}$
$\big[\text{Applying R}_3\rightarrow\text{R}_3-\text{R}_2\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2}\\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 5 & -2 & 2\end{bmatrix}\text{A}$
$\big[\text{Applying R}_3\rightarrow\ 2\text{R}_3\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1\rightarrow\text{R}_1+\frac{1}{2}\text{ R}_3\text{ and R}_2\rightarrow\text{R}_2-\frac{5}{2}\text{R}_3\Big]$
$\Rightarrow\text{A}^{-1}=\begin{bmatrix} 3 & -1 & 1\\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$

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