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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS4 Marks
Question
If $\text{A}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix},$ find the value of $\lambda$ so that $\text{A}^2=\lambda\text{A}-2\text{I}.$ Hence, find A-1.
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Answer

$\text{A}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$=\begin{bmatrix}1 & -2 \\4 & -4 \end{bmatrix}$
If $\text{A}^2=\lambda\text{A}-2\text{I}$
$\lambda\text{A}=\text{A}^2+2\text{I}$
$=\begin{bmatrix}1 & -2 \\4 & -4 \end{bmatrix}+\begin{bmatrix}2 & 0 \\0 & 2 \end{bmatrix}$
$\lambda\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$\lambda\begin{bmatrix}3\lambda & -2\lambda \\4\lambda & -2\lambda \end{bmatrix}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$3\lambda=3$
$\lambda=1$
$\text{A}^2=\text{A}-2\text{I}$
Px multiplying by A-1
A-1 AA = A-1 A - A-1 I
A = I - 2A-1
$2\text{A}^{-1}=\text{I}-\text{A}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}=\begin{bmatrix}-2 & 2 \\-4 & 3 \end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}-2 & 2 \\-4 & 3 \end{bmatrix}$

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