Mathematics - Matrices Question with Solution | TestHub

Given, $A^2=3A+\alpha I$

$\therefore A^3=A^2.A$

$\Rightarrow A^3=3A^2+\alpha A$

$\Rightarrow A^3=3\left(3A+\alpha I\right)+\alpha A$

$\Rightarrow A^3=9A+\alpha A+3\alpha I$

Now, $A^4=A^3.A$

$A^4=\left(9+\alpha \right)A^2+3\alpha A$

$=\left(9+\alpha \right)\left(3A+\alpha I\right)+3\alpha A$

$=\mathrm{A}\left(27+6\mathrm{\alpha }\right)+\alpha \left(9+\alpha \right).......\left(1\right)$

Given, $A^4=21A+\beta I.........\left(2\right)$

On comparing equation $\left(1\right)$ and equation $\left(2\right)$,

$\Rightarrow 27+6\alpha =21\Rightarrow \alpha =-1$

$\Rightarrow \beta =\alpha \left(9+\alpha \right)=-8$