Mathematics - Matrices Question with Solution | TestHub

Given,

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

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

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

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

Hence, $4\left(\alpha -\beta \right)=4\left(1\right)=4$