Step-by-Step Solution

1. Analyze the given condition: We are given the matrix $A=\left(\begin{array}{ccc} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{array}\right)$ and the expression $B = A^{7}-(\beta-1) A^{6}-\beta A^{5}$. The condition is that $B$ is a singular matrix. A matrix is singular if and only if its determinant is zero. So, we must have $\det(B) = 0$.

2. Simplify the matrix expression: We can factor out $A^5$ from the expression for $B$: $B = A^5(A^2 - (\beta-1)A - \beta I)$ where $I$ is the identity matrix.

3. Use the determinant property: Using the property $\det(XY) = \det(X)\det(Y)$, we have: $\det(B) = \det(A^5) \det(A^2 - (\beta-1)A - \beta I) = 0$ $\det(B) = (\det(A))^5 \det(A^2 - (\beta-1)A - \beta I) = 0$

4. Calculate the determinant of A: $\det(A) = \beta \begin{vmatrix} 1 & -2 \\ 1 & -2 \end{vmatrix} - 0 \begin{vmatrix} 2 & -2 \\ 3 & -2 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix}$ $\det(A) = \beta(1(-2) - (-2)(1)) - 0 + 1(2(1) - 3(1))$ $\det(A) = \beta(-2+2) + 1(2-3) = \beta(0) - 1 = -1$

5. Simplify the singularity condition: Since $\det(A) = -1$, $(\det(A))^5 = (-1)^5 = -1 \neq 0$. For $\det(B)$ to be zero, we must have: $\det(A^2 - (\beta-1)A - \beta I) = 0$

6. Relate the condition to eigenvalues: Let $\lambda$ be an eigenvalue of the matrix $A$. Then, the eigenvalues of the matrix polynomial $P(A) = A^2 - (\beta-1)A - \beta I$ are given by $P(\lambda) = \lambda^2 - (\beta-1)\lambda - \beta$. The determinant of a matrix is the product of its eigenvalues. For $\det(P(A))$ to be zero, at least one of its eigenvalues must be zero. Therefore, for some eigenvalue $\lambda$ of $A$, we must have: $\lambda^2 - (\beta-1)\lambda - \beta = 0$ Factoring this quadratic equation in $\lambda$: $(\lambda - \beta)(\lambda + 1) = 0$ This implies that an eigenvalue of $A$ must be either $\lambda = \beta$ or $\lambda = -1$.

7. Find the characteristic equation of A: The eigenvalues of $A$ are the roots of the characteristic equation $\det(A - \lambda I) = 0$. $A - \lambda I = \left(\begin{array}{ccc} \beta-\lambda & 0 & 1 \\ 2 & 1-\lambda & -2 \\ 3 & 1 & -2-\lambda \end{array}\right)$ $\det(A - \lambda I) = (\beta-\lambda)[(1-\lambda)(-2-\lambda) - (-2)(1)] - 0 + 1[2(1) - 3(1-\lambda)]$ $= (\beta-\lambda)[-2-\lambda+2\lambda+\lambda^2+2] + [2-3+3\lambda]$ $= (\beta-\lambda)(\lambda^2+\lambda) + (3\lambda-1)$ $= \beta\lambda^2 + \beta\lambda - \lambda^3 - \lambda^2 + 3\lambda - 1$ $= -\lambda^3 + (\beta-1)\lambda^2 + (\beta+3)\lambda - 1$ So, the characteristic equation is $\lambda^3 - (\beta-1)\lambda^2 - (\beta+3)\lambda + 1 = 0$.

8. Test the possible eigenvalues: We have two cases from step 6. Case 1: $\lambda = \beta$ is an eigenvalue of $A$. Substitute $\lambda = \beta$ into the characteristic equation: $\beta^3 - (\beta-1)\beta^2 - (\beta+3)\beta + 1 = 0$ $\beta^3 - (\beta^3 - \beta^2) - (\beta^2 + 3\beta) + 1 = 0$ $\beta^3 - \beta^3 + \beta^2 - \beta^2 - 3\beta + 1 = 0$ $-3\beta + 1 = 0 \implies \beta = \frac{1}{3}$

   Case 2: $\lambda = -1$ is an eigenvalue of $A$. Substitute $\lambda = -1$ into the characteristic equation: $(-1)^3 - (\beta-1)(-1)^2 - (\beta+3)(-1) + 1 = 0$ $-1 - (\beta-1)(1) + (\beta+3) + 1 = 0$ $-1 - \beta + 1 + \beta + 3 + 1 = 0$ $4 = 0$ This is a contradiction, so $\lambda = -1$ cannot be an eigenvalue.

9. Determine the value of $\beta$ and calculate the final answer: The only valid solution is $\beta = \frac{1}{3}$. The question asks for the value of $9\beta$. $9\beta = 9 \times \frac{1}{3} = 3$

   The final answer is 3.