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MATRICES — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMATRICES5 Marks
Question
If possible, using elementary row transformations, find the inverse of the following matrices:$\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}.$
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Answer

For getting the inverse of the given matrix A by row elementary operations we may write the given matrix as A = IA$\therefore\ \begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\text{A}$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\3&1&1\\0&1&3\end{bmatrix}\begin{bmatrix}1&0&0\\-1&1&0\\0&0&1\end{bmatrix}\text{A}\ [\because\ \text{R}_2\rightarrow\ \text{R}_2+\text{R}_1]$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\1&1&2\\2&1&2\end{bmatrix}\begin{bmatrix}1&0&0\\-2&1&0\\1&0&1\end{bmatrix}\text{A}\begin{bmatrix}\because\ \text{R}_2\rightarrow\ \text{R}_2+\text{R}_1\\\text{and }\text{R}_3\rightarrow\ \text{R}_3+\text{R}_1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\0&1&\frac{5}{2}\\4&1&1\end{bmatrix}\begin{bmatrix}1&0&0\\\frac{-5}{2}&1&0\\2&0&1\end{bmatrix}\text{A}\ \begin{bmatrix}\because\ \text{R}_3\rightarrow\ \text{R}_3+\text{R}_1\\\text{and }\text{R}_2\rightarrow\ \text{R}_2\frac{1}{2}\text{R}_1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\0&1&\frac{5}{2}\\0&1&3\end{bmatrix}\begin{bmatrix}1&0&0\\\frac{-5}{2}&1&0\\0&0&1\end{bmatrix}\text{A}\ [\because\ \text{R}_3\rightarrow\ \text{R}_3-2\text{R}_1]$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\0&1&\frac{5}{2}\\0&0&\frac{1}{2}\end{bmatrix}\begin{bmatrix}1&0&0\\\frac{-5}{2}&1&0\\\frac{5}{2}&-1&1\end{bmatrix}\text{A}\ [\because\ \text{R}_3\rightarrow\ \text{R}_3-\text{R}_2]$
$\Rightarrow\ \begin{bmatrix}1&0&\frac{-1}{2}\\0&1&\frac{5}{2}\\0&0&1\end{bmatrix}\begin{bmatrix}\frac{1}{2}&0&0\\\frac{-5}{2}&1&0\\5&-2&2\end{bmatrix}\text{A}\ \begin{bmatrix}\because\ \text{R}_1\rightarrow\ \frac{1}{2}\text{R}_1\\\text{and }\text{R}_ 3\rightarrow\ 2\text{R}_3\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\text{A}\ \begin{bmatrix}\because\ \text{R}_1\rightarrow\ \text{R}_1\frac{1}{2}\text{R}_3\\\text{and }\text{R}_2\rightarrow\ \text{R}_2-\frac{5}{2}\text{R}_3\end{bmatrix}$
Hence, $\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$ is the inverse of given matrix.

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