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

The correct answer is 3.

We are given the matrix:

$A=\left[\begin{array}{ccc}\beta & 0& 1\\ 2& 1& -2\\ 3& 1& -2\end{array}\right]$

We are also given that the matrix expression $M=A^7-(\beta -1)A^6-\beta A^5$ is a singular matrix.

A matrix is singular if and only if its determinant is equal to zero. Therefore:

$det\left(M\right)=0$

Let's factor out $A^5$ from the expression $M$:

$M=A^5(A^2-(\beta -1)A-\beta I)$

We can factor the quadratic polynomial inside the parentheses as:

$A^2-(\beta -1)A-\beta I=(A-\beta I)(A+I)$

Thus, the matrix expression becomes:

$M=A^5(A-\beta I)(A+I)$

Taking the determinant on both sides:

$det\left(M\right)=det\left(A^5\right)·det(A-\beta I)·det(A+I)=0$

First, let's find the determinant of matrix $A$:

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

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

Since $det\left(A\right)=-1\ne 0$, we have $det\left(A^5\right)=(-1{)}^5=-1\ne 0$.

Next, let's check $det(A+I)$:

$A+I=\left[\begin{array}{ccc}\beta +1& 0& 1\\ 2& 2& -2\\ 3& 1& -1\end{array}\right]$

$det(A+I)=(\beta +1)\left(2\right(-1)-(-2\left)\right(1\left)\right)+1\left(2\right(1)-3(2\left)\right)$

$det(A+I)=(\beta +1)\left(0\right)+1(2-6)=-4\ne 0$

Since both $det\left(A^5\right)\ne 0$ and $det(A+I)\ne 0$, for $det\left(M\right)=0$ to hold, we must have:

$det(A-\beta I)=0$

This implies that $\beta$ must be an eigenvalue of matrix $A$.

Now, let's write out the characteristic matrix $A-\beta I$:

$A-\beta I=\left[\begin{array}{ccc}0& 0& 1\\ 2& 1-\beta & -2\\ 3& 1& -2-\beta \end{array}\right]$

Evaluating its determinant along the first row gives:

$det(A-\beta I)=1·\left(2\right(1)-3(1-\beta \left)\right)=0$

Simplifying the linear equation in terms of $\beta$:

$2-3+3\beta =0$

$3\beta -1=0\Rightarrow \beta =\frac{1}{3}$

Finally, we calculate the required value of $9\beta$:

$9\beta =9·\frac{1}{3}=3$