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(B) Given $A = \begin{bmatrix} \beta & 0 & 1 \ 2 & 1 & -2 \ 3 & 1 & -2 \end{bmatrix}$.First,calculate the determinant $|A| = \beta(1(-2) - (-2)(1))

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$3$

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(B) Given $A = \begin{bmatrix} \beta & 0 & 1 \ 2 & 1 & -2 \ 3 & 1 & -2 \end{bmatrix}$.First,calculate the determinant $|A| = \beta(1(-2) - (-2)(1)) - 0 + 1(2(1) - 3(1)) = \beta(0) + 1(-1) = -1$.Since $|A| = -1 
eq 0$,the matrix $A$ is invertible.We are given that $A^7 - (\beta - 1)A^6 - \beta A^5$ is a singular matrix,so its determinant is $0$:$|A^5(A^2 - (\beta - 1)A - \beta I)| = 0$.Since $|A| 
eq 0$,we have $|A|^5 |A^2 - (\beta - 1)A - \beta I| = 0$,which implies $|A^2 - (\beta - 1)A - \beta I| = 0$.Factor the expression inside the determinant:$A^2 - (\beta - 1)A - \beta I = A^2 - \beta A + A - \beta I = A(A - \beta I) + I(A - \beta I) = (A + I)(A - \beta I)$.Thus,$|A + I| |A - \beta I| = 0$.Calculate $A + I = \begin{bmatrix} \beta + 1 & 0 & 1 \ 2 & 2 & -2 \ 3 & 1 & -1 \end{bmatrix}$.$|A + I| = (\beta + 1)(-2 - (-2)) - 0 + 1(2 - 6) = (\beta + 1)(0) - 4 = -4 
eq 0$.Therefore,we must have $|A - \beta I| = 0$.$A - \beta I = \begin{bmatrix} 0 & 0 & 1 \ 2 & 1 - \beta & -2 \ 3 & 1 & -2 - \beta \end{bmatrix}$.$|A - \beta I| = 1(2 - 3(1 - \beta)) = 2 - 3 + 3\beta = 3\beta - 1$.Setting $3\beta - 1 = 0$,we get $\beta = \frac{1}{3}$.Thus,$9\beta = 9 	imes \frac{1}{3} = 3$.

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

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