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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 Marks
Question
Using elementary transformations, find the inverse of the matrix
$\begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -1\\ 2 & 1 & 0 \end{pmatrix}.$
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Answer

Given matrix A can be written as
$\begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\text{A}$
$\text{Applying R}_{2}\rightarrow\text{R}_{2}+\text{3R}_{1}\text{ R}_{3}\rightarrow{\text{R}_{3}-\text{2R}_{1}}$ $\begin{pmatrix} 1 & 3 & -2 \\ 0 & 9 & -7 \\ 0 & -5 & 4 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2}+\text{2R}_{3}\Rightarrow\begin{pmatrix} 1 & 3 & -2 \\ 0 & -1& 1 \\ 0 & -5 & 4 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 2 \\ -2 & 0 & 1 \end{pmatrix}\text{A}$
$\text{R}_{3}\rightarrow\text{R}_{3}-\text{5R}_{2}\Rightarrow\begin{pmatrix} 1 & 3 & -2 \\ 0 & -1& 1 \\ 0 & 0 & -1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 2 \\ 3 & -5 & -9 \end{pmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2}+\text{R}_{3},\text{R}_{3}\rightarrow\begin{pmatrix} 1 & 3 & -2 \\ 0 & -1& 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 2 & -4 & -7 \\ -3 & 5 & 9 \end{pmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1}+\text{2R}_{3},\text{R}_{2}\rightarrow\text{-R}_{3}\begin{pmatrix} 1 & 3 & 0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} -5 & 10 & 18 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{pmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1}-\text{3R}_{2}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{pmatrix}\text{A}$
$\therefore\text{A}^{-1}\begin{pmatrix} 1 & -2 & -3 \\ -2 & 4& 7 \\ -3 & 5 & 9 \end{pmatrix}.$

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