MCQ
Let $A\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right] .$ If $A^{-1}=\alpha I+\beta A, \alpha, \beta \in R, I$ is a $2 \times 2$ identity matrix, then $4(\alpha-\beta)$ is equal to:
- A$5$
- ✓$4$
- C$2$
- D$\frac{8}{3}$
✓
Answer
Correct option: B.
$4$
b
$A=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right],|A|=6$
$A=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right],|A|=6$
$\mathrm{A}^{-1}=\frac{\operatorname{adjA}}{|\mathrm{A}|}=\frac{1}{6}\left[\begin{array}{cc}4 & -2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]$
$\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]=\left[\begin{array}{ll}\alpha & 0 \\ 0 & \alpha\end{array}\right]+\left[\begin{array}{cc}\beta & 2 \beta \\ -\beta & 4 \beta\end{array}\right]$
$\alpha+\beta=\frac{2}{3}$
$\beta=-\frac{1}{6}$
$\Rightarrow \alpha=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$
$4(\alpha-\beta)=4(1)=4$
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