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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Find the inverse of the following matrices by using elementry row transformation:
$\begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix}$
✓

Answer

$\text{A}=\begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix}$
We have A = IA
$\Rightarrow\begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 2 & 0 \\ 0 & -1 & -1 \\ 0 & -3 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-2\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-\text{R}_1\big]$
$\Rightarrow\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & -3 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & -1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow-\text{R}_2\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 6 \end{bmatrix}=\begin{bmatrix} -3 & 2 & 0 \\ 2 & -1 & 0 \\ 5 & -3 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1-2\text{R}_2\text{ and R}_3\rightarrow\text{R}_3+3\text{R}_2\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -3 & 2 & 0 \\ 2 & -1 & 0 \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_3\rightarrow\frac{1}{6}\text{ R}_3\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{3} & -\frac{1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{bmatrix}\text{A}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1+2\text{R}_1\text{ and R}_2\rightarrow\text{R}_2-\text{R}_3\big]$
$\therefore\ \text{A}^{-1}=\begin{bmatrix} -\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{3} & -\frac{1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{bmatrix}$

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