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Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
Question
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
✓

Answer

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

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