Singularity Condition for Matrix Polynomial Equation

To find the value of $9\beta$, we analyze the given matrix polynomial equation and use the properties of matrix determinants.

The given matrix is: $A = \begin{pmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{pmatrix}$

We are given that the matrix $M = A^7 - (\beta - 1)A^6 - \beta A^5$ is singular, which means its determinant is equal to zero: $\det(M) = 0$

First, we factor out $A^5$ from the matrix expression: $M = A^5 \left( A^2 - (\beta - 1)A - \beta I \right)$

The quadratic term inside the parentheses can be factored as: $A^2 - (\beta - 1)A - \beta I = (A - \beta I)(A + I)$

Thus, we can write $M$ as: $M = A^5 (A - \beta I)(A + I)$

Taking the determinant on both sides: $\det(M) = \det(A^5) \cdot \det(A - \beta I) \cdot \det(A + I) = 0$ $\implies (\det(A))^5 \cdot \det(A - \beta I) \cdot \det(A + I) = 0$

Now, let's evaluate each factor's determinant:

1. Determinant of $A$: $\det(A) = \begin{vmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{vmatrix}$ Expanding along the first row: $\det(A) = \beta \left( 1(-2) - (-2)(1) \right) - 0 + 1 \left( 2(1) - 3(1) \right)$ $\det(A) = \beta(0) + 1(-1) = -1 \neq 0$ Therefore, $(\det(A))^5 = (-1)^5 = -1 \neq 0$.

2. Determinant of $A + I$: $A + I = \begin{pmatrix} \beta + 1 & 0 & 1 \\ 2 & 2 & -2 \\ 3 & 1 & -1 \end{pmatrix}$ Expanding along the first row: $\det(A + I) = (\beta + 1) \left( 2(-1) - (-2)(1) \right) - 0 + 1 \left( 2(1) - 3(2) \right)$ $\det(A + I) = (\beta + 1)(0) + 1(2 - 6) = -4 \neq 0$

3. Determinant of $A - \beta I$: Since $(\det(A))^5 \neq 0$ and $\det(A + I) \neq 0$, for $\det(M) = 0$, we must have: $\det(A - \beta I) = 0$

   Evaluating $\det(A - \beta I)$: $A - \beta I = \begin{pmatrix} 0 & 0 & 1 \\ 2 & 1 - \beta & -2 \\ 3 & 1 & -2 - \beta \end{pmatrix}$ Expanding along the first row: $\det(A - \beta I) = 1 \cdot \left[ 2(1) - 3(1 - \beta) \right] = 2 - 3 + 3\beta = 3\beta - 1$

Setting the determinant to zero gives: $3\beta - 1 = 0 \implies \beta = \frac{1}{3}$

Finally, we find the value of $9\beta$: $9\beta = 9 \times \frac{1}{3} = 3$