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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Using elementary transformations, find the inverse of the following matrix:
$\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix} $
✓

Answer

$\text{A = I A}$
$\Rightarrow \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \text{A} $
$\text{R}_{1}\leftrightarrow \text{R}_{2}$
$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\text{A}$
$\text{R}_{3}\rightarrow\text{R}_{3} - \text{3 R}_{1}$
$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -9 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{bmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1} - \text{2 R}_{2},\ \ \text{R}_{3} \rightarrow\text{R}_{3} + \text{5 R}_{2}$
$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{bmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1} + \text{ R}_{3}, \text{R}_{2}\rightarrow\text{R}_{2} - \text{2 R}_{3}$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -2 & 1 \\ -9 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix}\text{A}$ [operating Row wise to reach at this step]
$\text{A}^{-1} = \begin{bmatrix} 3 & -2 & 1 \\ -9 & 6 & -2\\ 5 & -3 & 1 \end{bmatrix}$

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