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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Find the inverse of the following matrix using elementary operations. $\text{A}=\begin{bmatrix}1 & 2 & -2\\-1 & 3 & 0\\0 & -2 & 1 \end{bmatrix}$
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Answer

We know that
$\text{A}^{-1} = \text{I}^{}\text{A}$
Or, $\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\-1 & 3 & 0\\0 & -2 & 1 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\0 & 5 & -2\\0 & -2 & 1 \end{bmatrix}$ [Applying $R_2 → R_2 → R_1$]
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 2\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\0 & 1 & 0\\0 & -2 & 1 \end{bmatrix}$ [Applying $R_2 → R_2 → R_3$]
$\text{A}^{-1}=\begin{bmatrix}-1 & -2 & -4\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & -2\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$ [Applying $R_1 → R_1 + (-2)R_2, R_3 → R_3 + 2R_2$]
$\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$ [Applying $R_2 → R_2 → R_3$]
Hence, $\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}$

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