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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using elementary transformations, find the inverse of the following matrix:
$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix}$
✓

Answer

$\text{A}=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix}=\begin{pmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_2 \rightarrow R_2 - 2R_1, R_3 \rightarrow R_3 + 2R_1$, we get
$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 &0 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_1 \rightarrow R_1 - 2R_2$, we get
$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 5 & -2 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_1 \rightarrow R_1 - R_3$, we get
$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 3 & -2 & -1 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_2 \rightarrow R_2 - R_3$, we get
$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
$\therefore \text{ A}^{-1}=\begin{pmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix}.$

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