Let β be a real number. Consider the matrix A = ( cases β & 0 & 1 2...

Given,

A = \left(\begin{cases}\beta & 0 & 1 \\ 2 & 1 & - 2 \\ 3 & 1 & - 2\end{cases}\right)

So, $det \left(A\right) = - 1 . . . \left(1\right)$

Now for ${A}^{7} - \left(\beta - 1\right){A}^{6} - \beta{A}^{5} to be singular</p><p>\left|{A}^{7} - \left(\beta - 1\right){A}^{6} - \beta{A}^{5}\right| = 0$

$⇒ \left|{A}^{5}\right|\left|{A}^{2} - ( \beta - 1 ) A - \beta\right| = 0 . . . \left(2\right)$

$⇒ \left|{A}^{5}\right|\left|\left(A + I\right)\left(A - \betal\right)\right| = 0$

$∴ \left|{A}^{5}\right|\left|A + I\right|\left|A - \beta\right| = 0$

As \left|A\right| \ne 0$</p><p>\left|A + 1\right|$ or \left|A - \beta\right| = 0$</p><p>$⇒ \left($\begin{cases}\beta + 1 & 0 & 1 \\ 2 & 2 & - 2 \\ 3 & 1 & - 1\end{cases}$\right) = 0 ⇒ \left|A + I\right| = - 4, \text{here}\left|A\right| \ne 0 \text{ and } \left|A + I\right| \ne 0</p><p>Now solving \left|A - \betal\right| = \left|\begin{cases}0 & 0 & 1 \\ 2 & 1 - \beta & - 2 \\ 3 & 1 & - 2 - \beta\end{cases}\right| = 0

$⇒ 2 - 3\left(1 - \beta\right) = 0$ $⇒ 2 - 3 + 3\beta = 0$

$⇒ \beta = \dfrac{1}{3}$

$∴ 9\beta = 3$