This problem requires a strong understanding of matrix properties, particularly those involving cofactors, adjoints, and determinants. We will leverage the relationships between these concepts to solve for unknown variables and ultimately determine the required determinant.

Key Concepts and Formulas

1. Cofactor Matrix: For a matrix $A = [a_{ij}]$, the cofactor matrix, let's call it $C$, has elements $C_{ij}$ where $C_{ij} = (-1)^{i+j} M_{ij}$. $M_{ij}$ is the minor of $a_{ij}$ (the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column).
2. Adjoint Matrix: The adjoint of a matrix $A$, denoted as $\operatorname{adj}(A)$, is the transpose of its cofactor matrix. That is, $\operatorname{adj}(A) = C^T$.
3. Fundamental Matrix Property: For any square matrix $A$, $A \cdot \operatorname{adj}(A) = \operatorname{det}(A) \cdot I$, where $I$ is the identity matrix of the same order as $A$.
4. Determinant of a Product: For two square matrices $X$ and $Y$ of the same order, $\operatorname{det}(XY) = \operatorname{det}(X) \cdot \operatorname{det}(Y)$.
5. Determinant of a Transpose: For any square matrix $M$, $\operatorname{det}(M^T) = \operatorname{det}(M)$.
6. Determinant of a Scalar Multiple of Identity: For an $n \times n$ identity matrix $I$ and a scalar $k$, $\operatorname{det}(kI) = k^n$.
7. Determinant of Cofactor Matrix: If $C$ is the cofactor matrix of an $n \times n$ matrix $A$, then $\operatorname{det}(C) = (\operatorname{det}(A))^{n-1}$. This follows directly from $A \cdot C^T = \operatorname{det}(A) \cdot I$, taking determinants on both sides: $\operatorname{det}(A) \cdot \operatorname{det}(C^T) = (\operatorname{det}(A))^n \Rightarrow \operatorname{det}(A) \cdot \operatorname{det}(C) = (\operatorname{det}(A))^n$. If $\operatorname{det}(A) \neq 0$, then $\operatorname{det}(C) = (\operatorname{det}(A))^{n-1}$.

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Step 1: Understand the Given Information and Establish Key Relationships

We are given two $3 \times 3$ matrices: $A=\left[\begin{array}{rrr}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha\end{array}\right]$ $B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta\end{array}\right]$ We are told that $B$ is the matrix of cofactors of the elements of $A$. This means that each element $B_{ij}$ in matrix $B$ is equal to the cofactor $C_{ij}$ of the corresponding element $a_{ij}$ in matrix $A$. Therefore, $B$ is the cofactor matrix of $A$.

Our goal is to find $\operatorname{det}(AB)$. Using the property of determinant of a product, $\operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B)$. So, we need to find $\operatorname{det}(A)$ and $\operatorname{det}(B)$.

Step 2: Determine the Values of $\alpha$ and $\beta$

To find $\operatorname{det}(A)$ and $\operatorname{det}(B)$, we first need to determine the specific values of $\alpha$ and $\beta$. We can do this by calculating selected cofactors of $A$ and equating them to the corresponding elements in $B$. We are given that $\alpha \beta \neq 0$.

Let's calculate a few cofactors of $A$:

1. Cofactor $C_{31}$ (for element $a_{31}=-\beta$): $C_{31} = (-1)^{3+1} \det \left[\begin{array}{cc}\alpha & 3 \\ \alpha & \beta\end{array}\right] = 1 \cdot (\alpha \cdot \beta - 3 \cdot \alpha) = \alpha \beta - 3 \alpha$ From matrix $B$, the element $B_{31} = -2\alpha$. Equating $C_{31}$ with $B_{31}$: $\alpha \beta - 3 \alpha = -2 \alpha$ $\alpha \beta - \alpha = 0$ $\alpha (\beta - 1) = 0$ Since it is given that $\alpha \beta \neq 0$, it implies $\alpha \neq 0$. Therefore, we must have $\beta - 1 = 0$, which gives us $\boxed{\beta = 1}$.

2. Cofactor $C_{32}$ (for element $a_{32}=\alpha$): $C_{32} = (-1)^{3+2} \det \left[\begin{array}{cc}\beta & 3 \\ \alpha & \beta\end{array}\right] = -1 \cdot (\beta \cdot \beta - 3 \cdot \alpha) = -(\beta^2 - 3\alpha)$ From matrix $B$, the element $B_{32} = 5$. Equating $C_{32}$ with $B_{32}$: $-(\beta^2 - 3\alpha) = 5$ $\beta^2 - 3\alpha = -5$ Now, substitute the value $\beta = 1$ into this equation: $(1)^2 - 3\alpha = -5$ $1 - 3\alpha = -5$ $-3\alpha = -6$ $\boxed{\alpha = 2}$

We have found $\alpha=2$ and $\beta=1$. Self-check: Let's verify these values with one more cofactor comparison, for example, $C_{11}$. $C_{11} = (-1)^{1+1} \det \left[\begin{array}{cc}\alpha & \beta \\ \alpha & 2 \alpha\end{array}\right] = 1 \cdot (\alpha \cdot 2\alpha - \beta \cdot \alpha) = 2\alpha^2 - \alpha\beta$ From matrix $B$, the element $B_{11} = 3\alpha$. Equating $C_{11}$ with $B_{11}$: $2\alpha^2 - \alpha\beta = 3\alpha$ Substitute $\alpha=2$ and $\beta=1$: $2(2)^2 - (2)(1) = 3(2)$ $2(4) - 2 = 6$ $8 - 2 = 6$ $6 = 6$ The values $\alpha=2$ and $\beta=1$ are consistent.

Step 3: Calculate $\operatorname{det}(A)$

Now that we have $\alpha=2$ and $\beta=1$, we can write matrix $A$ explicitly: $A=\left[\begin{array}{rrr}1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4\end{array}\right]$ Let's calculate its determinant using cofactor expansion along the first row: $\operatorname{det}(A) = 1 \cdot \det \left[\begin{array}{cc}2 & 1 \\ 2 & 4\end{array}\right] - 2 \cdot \det \left[\begin{array}{cc}2 & 1 \\ -1 & 4\end{array}\right] + 3 \cdot \det \left[\begin{array}{cc}2 & 2 \\ -1 & 2\end{array}\right]$ $= 1 \cdot (2 \cdot 4 - 1 \cdot 2) - 2 \cdot (2 \cdot 4 - 1 \cdot (-1)) + 3 \cdot (2 \cdot 2 - 2 \cdot (-1))$ $= 1 \cdot (8 - 2) - 2 \cdot (8 + 1) + 3 \cdot (4 + 2)$ $= 1 \cdot (6) - 2 \cdot (9) + 3 \cdot (6)$ $= 6 - 18 + 18$ $\operatorname{det}(A) = 6$

Step 4: Calculate $\operatorname{det}(B)$

We know that $B$ is the cofactor matrix of $A$. For an $n \times n$ matrix, the determinant of its cofactor matrix is given by $\operatorname{det}(B) = (\operatorname{det}(A))^{n-1}$. In this case, $n=3$. So, $\operatorname{det}(B) = (\operatorname{det}(A))^{3-1} = (\operatorname{det}(A))^2$. Using the value $\operatorname{det}(A) = 6$: $\operatorname{det}(B) = (6)^2 = 36$

Step 5: Calculate $\operatorname{det}(AB)$

Finally, we use the property of the determinant of a product: $\operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B)$ Substitute the values we found: $\operatorname{det}(AB) = 6 \cdot 36$ $\operatorname{det}(AB) = 216$

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Tips for Success / Common Pitfalls

• Cofactor vs. Adjoint: Remember that the cofactor matrix is not the adjoint matrix. The adjoint matrix is the transpose of the cofactor matrix. $\operatorname{adj}(A) = B^T$.
• Determinant Properties: Be familiar with the properties of determinants, especially $\operatorname{det}(XY) = \operatorname{det}(X)\operatorname{det}(Y)$ and $\operatorname{det}(kI) = k^n$.
• Sign Errors: When calculating cofactors, be careful with the $(-1)^{i+j}$ term.
• Condition $\alpha \beta \neq 0$: Always pay attention to given conditions. This condition was crucial for deducing $\beta-1=0$ from $\alpha(\beta-1)=0$.
• Verification: After finding $\alpha$ and $\beta$, it's good practice to verify them with another cofactor calculation or by checking if $\operatorname{det}(A) \neq 0$ (which it isn't, $6 \neq 0$).

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Summary / Key Takeaway

This problem effectively tests your understanding of the definitions of cofactor and adjoint matrices, and their deep connection to the determinant of a matrix. The key steps involved:

1. Using the definition of the cofactor matrix to establish equations for unknown variables.
2. Solving for these variables.
3. Calculating the determinant of the original matrix.
4. Applying the property $\operatorname{det}(\text{Cofactor Matrix of A}) = (\operatorname{det}(A))^{n-1}$.
5. Finally, using the determinant product rule $\operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B)$.

The final answer is $\boxed{\text{216}}$.