Solve this

`
Question:

Let $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right] .$ If $\mathrm{A}^{-1}=\alpha \mathrm{I}+\beta \mathrm{A}, \alpha, \beta \in \mathbf{R}, \mathrm{I}$ is a $2 \times 2$ identity matrix, then $4(\alpha-\beta)$ is equal to:

  1. 5

  2. $\frac{8}{3}$

  3. 2

  4. 4


Correct Option: , 4

Solution:

$\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right],|\mathrm{A}|=6$

$\mathrm{A}^{-1}=\frac{\mathrm{adj} \mathrm{A}}{|\mathrm{A}|}=\frac{1}{6}\left[\begin{array}{cc}4 & -2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]$

$\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]=\left[\begin{array}{cc}\alpha & 0 \\ 0 & \alpha\end{array}\right]+\left[\begin{array}{cc}\beta & 2 \beta \\ -\beta & 4 \beta\end{array}\right]$

$\left.\begin{array}{l}\alpha+\beta=\frac{2}{3} \\ \beta=-\frac{1}{6}\end{array}\right\} \Rightarrow \alpha=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$]

$4(\alpha-\beta)=4(1)=4$

Leave a comment