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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Using elementary row operations, find the inverse of the following matrix:
$\text{A} = \begin{pmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{pmatrix}$
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Answer

$\text{A = IA}\therefore \begin{pmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2} -2\text{R}_{3}\Rightarrow \begin{pmatrix} 2 & -1 & 3 \\ 1 & -1 & -5 \\ -3 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}\text{A}$
$\begin{matrix} \text{R}_{1}\rightarrow\text{R}_{1} -2\text{R}_{2} \\ \text{R} {3}\rightarrow\text{R}_{3} +3\text{R}_{2} \\ \end{matrix}\Rightarrow \begin{pmatrix} 0 & 1 & 13 \\ 1 & -1 & -5 \\ 0 & -1 & -12 \end{pmatrix} = \begin{pmatrix} 1 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 3 & -5 \end{pmatrix}\text{A} $
$\text{R}_{1}\leftrightarrow\text{R}_{2} \Rightarrow \begin{pmatrix} 1 & -1 & -5 \\ 0 & 1 & 13 \\ 0 & -1 & -12 \end{pmatrix} = \begin{pmatrix} 0 & 1 & -2 \\ 1 & -2 & 4 \\ 0 & 3 & -5 \end{pmatrix}\text{A} $
$ \begin{matrix} \text{R}_{1}\rightarrow\text{R}_{1} + \text{R}_{2} \\ \text{R}_{3}\rightarrow\text{R}_{3} + \text{R}_{2} \\ \end{matrix}\Rightarrow \begin{pmatrix} 1 & 0 & 8 \\ 0 & 1 & 13 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 2 \\ 1 & -2 & 4 \\ 1 & 1 & -5 \end{pmatrix}\text{A} $
$ \begin{matrix} \text{R}_{1}\rightarrow\text{R}_{1} - \text{8R}_{3} \\ \text{R}_{2}\rightarrow\text{R}_{2} - \text{13R}_{3} \\ \end{matrix}\Rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{pmatrix}\text{A} $
$\Rightarrow\text{A}^{-1} = \begin{pmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{pmatrix}$

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