Question

Let $\beta$ be a real number. Consider the matrix $A=\begin{pmatrix}\beta & 0 & 1 \\2 & 1 & -2 \\3 & 1 & -2\end{pmatrix}$ If $A^7-(\beta-1) A^6-\beta A^5$ is a singular matrix, then the value of $9 \beta$ is _____

Answer 1

$ A=\begin{pmatrix}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{pmatrix}|A|=-1 $
$ \Rightarrow\left|A^7-(\beta-1) A^6-\beta A^5\right|=0 $
$\Rightarrow|A|^5\left|A^2-(\beta-1) A-\beta I\right|=0$
$\Rightarrow|A|^5\left|\left(A^2-\beta A \right)+A-\beta I\right|=0$
$\Rightarrow|A|^5|A(A-\beta I)+I(A-\beta I)|=0$
$|A|^5|(A+I)(A-\beta I)|=0$
$ A+I=\begin{pmatrix}\beta+1 & 0 & 1 \\ 2 & 2 & -2 \\ 3 & 1 & -1\end{pmatrix} \Rightarrow|A+I|=-4$,
Here $| A | \neq 0 \&|A+I| \neq 0 $
$ A-\beta I=\begin{pmatrix}0 & 0 & 1 \\ 2 & 1-\beta & -2 \\ 3 & 1 & -2-\beta\end{pmatrix} $
$ |A-\beta I|=2-3(1-\beta)=3 \beta-1=0 \Rightarrow \beta=\frac{1}{3} $
$ 9 \beta=3$