Key Concepts and Formulas

This problem relies on fundamental properties of determinants and adjoints of matrices. For an $n \times n$ matrix $X$ and a scalar $k$:

1. Determinant of a scalar multiple of a matrix: $\det(kX) = k^n \det(X)$ This property states that multiplying every element of a matrix by a scalar $k$ scales its determinant by $k^n$, where $n$ is the order of the matrix.

2. Determinant of the adjoint of a matrix: $\det(\operatorname{adj}(X)) = (\det(X))^{n-1}$ The determinant of the adjoint of a matrix $X$ is equal to the determinant of $X$ raised to the power of $(n-1)$.

In this problem, the matrix $A$ is of order $3 \times 3$, so $n=3$. Therefore, the formulas relevant to this problem become:

• $\det(kX) = k^3 \det(X)$
• $\det(\operatorname{adj}(X)) = (\det(X))^2$

Step-by-Step Solution

Step 1: Determine the value of 'a'.

Our first goal is to find the value of the real number 'a'. We are given the matrix $A+I$ and the determinant of $A$. We can use these to form an equation for 'a'.

1. Express matrix $A$ in terms of $A+I$ and $I$: We know that $A = (A+I) - I$. Given $A+I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$ and the identity matrix $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Subtracting $I$ from $A+I$: $A = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1-1 & a-0 & 1-0 \\ 2-0 & 1-1 & 0-0 \\ a-0 & 1-0 & 2-1 \end{bmatrix} = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}$

2. Calculate $\det(A)$ in terms of 'a': We will calculate the determinant of matrix $A$ using cofactor expansion along the second row, as it contains two zeros, simplifying the calculation. $\det(A) = -2 \cdot \det\begin{pmatrix} a & 1 \\ 1 & 1 \end{pmatrix} + 0 \cdot \det\begin{pmatrix} 0 & 1 \\ a & 1 \end{pmatrix} - 0 \cdot \det\begin{pmatrix} 0 & a \\ a & 1 \end{pmatrix}$ $\det(A) = -2(a \cdot 1 - 1 \cdot 1) + 0 - 0$ $\det(A) = -2(a - 1) = -2a + 2$

3. Equate $\det(A)$ to the given value to find 'a': We are given that $\det(A) = -4$. $-2a + 2 = -4$ $-2a = -4 - 2$ $-2a = -6$ $a = 3$ Thus, the value of 'a' is 3.

Step 2: Substitute the value of 'a' into the target expression.

The expression we need to evaluate is $\det((a+1) \operatorname{adj}((a-1) A))$. Substitute $a=3$ into this expression: $\det((3+1) \operatorname{adj}((3-1) A)) = \det(4 \operatorname{adj}(2A))$ This is the simplified expression we need to work with.

Step 3: Apply determinant properties iteratively.

We will apply the key determinant properties step-by-step, working from the outermost determinant inwards. Remember $n=3$ for all matrices involved.

1. Apply $\det(kX) = k^3 \det(X)$ for the outermost determinant: Here, $k=4$ and $X = \operatorname{adj}(2A)$. $\det(4 \operatorname{adj}(2A)) = 4^3 \det(\operatorname{adj}(2A))$

2. Apply $\det(\operatorname{adj}(Y)) = (\det(Y))^2$ for the adjoint part: Here, $Y=2A$. $4^3 \det(\operatorname{adj}(2A)) = 4^3 (\det(2A))^2$

3. Apply $\det(kX) = k^3 \det(X)$ for the innermost determinant: Here, $k=2$ and $X=A$. $\det(2A) = 2^3 \det(A)$ Substitute this back into the expression: $4^3 (\det(2A))^2 = 4^3 (2^3 \det(A))^2$

4. Simplify the powers: Express $4$ as $2^2$: $(2^2)^3 \cdot (2^3)^2 \cdot (\det(A))^2$ Using the power rule $(x^p)^q = x^{pq}$: $2^{2 \cdot 3} \cdot 2^{3 \cdot 2} \cdot (\det(A))^2$ $2^6 \cdot 2^6 \cdot (\det(A))^2$ Using the power rule $x^p \cdot x^q = x^{p+q}$: $2^{6+6} \cdot (\det(A))^2 = 2^{12} (\det(A))^2$

Step 4: Substitute the value of $\det(A)$ and simplify.

We are given $\det(A) = -4$. Substitute this value into our simplified expression: $2^{12} (\det(A))^2 = 2^{12} (-4)^2$ $= 2^{12} (16)$ Since $16 = 2^4$: $= 2^{12} \cdot 2^4$ $= 2^{12+4}$ $= 2^{16}$

Step 5: Determine m and n, then calculate m+n.

The problem states that the value of the expression is $2^m 3^n$. We found the value to be $2^{16}$. To match the format $2^m 3^n$, we can write $2^{16}$ as $2^{16} \cdot 3^0$. Comparing this with $2^m 3^n$: $m = 16$ $n = 0$

Both $m=16$ and $n=0$ are within the specified range $\{0, 1, 2, \ldots, 20\}$. Finally, we need to find $m+n$: $m+n = 16 + 0 = 16$

Common Mistakes & Tips

• Incorrect Order of Matrix (n): A common error is to use an incorrect value for 'n' (the order of the matrix) in the determinant properties. Always double-check that $n=3$ for $3 \times 3$ matrices.
• Misapplication of Scalar Multiplication: Remember that $\det(kX) = k^n \det(X)$, not $k \det(X)$. Also, when applying this to $\det(k \operatorname{adj}(M))$, the scalar $k$ is raised to the power of $n$ because $\operatorname{adj}(M)$ is also an $n \times n$ matrix.
• Algebraic Errors in Determinant Calculation: Be careful with signs and calculations when expanding the determinant of a $3 \times 3$ matrix.
• Simplification of Powers: Ensure accurate combination of powers, especially when dealing with negative bases like $(-4)^2 = 16$.

Summary

We first determined the value of 'a' by calculating $\det(A)$ from the given $A+I$ matrix and equating it to the provided $\det(A)=-4$. With $a=3$, we substituted this into the complex expression $\det((a+1) \operatorname{adj}((a-1) A))$. We then systematically applied the determinant properties for scalar multiplication ($\det(kX) = k^3 \det(X)$) and adjoints ($\det(\operatorname{adj}(X)) = (\det(X))^2$) multiple times. This process simplified the expression to $2^{16}$. By comparing this to the given format $2^m 3^n$, we identified $m=16$ and $n=0$, leading to $m+n=16$.

The final answer is $\boxed{\text{16}}$, which corresponds to option (D).