Let α and β be real numbers. Consider a 3 × 3 matrix A such that A...

Given, ${A}^{2} = 3A + \alphaI$

$∴ {A}^{3} = {A}^{2} . A$

$⇒ {A}^{3} = 3{A}^{2} + \alphaA$

$⇒ {A}^{3} = 3\left(3A + \alphaI\right) + \alphaA$

$⇒ {A}^{3} = 9A + \alphaA + 3\alphaI$

Now, ${A}^{4} = {A}^{3} . A$

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

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

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

Given, ${A}^{4} = 21A + \betaI . . . . . . . . . \left(2\right)</p><p>On comparing equation \left(1\right)$ and equation \left(2\right),

$⇒ 27 + 6\alpha = 21 ⇒ \alpha = - 1$

$⇒ \beta = \alpha\left(9 + \alpha\right)$ = - 8$</p>$