Key Concepts and Formulas

Before we dive into the solution, let's recall the fundamental properties of determinants and adjoints that are crucial for solving this problem. For an $N \times N$ matrix $A$:

1. Determinant of a scalar multiple of a matrix: If $k$ is a scalar, then $|kA| = k^N |A|$. Here, $N$ represents the order (dimension) of the matrix. For a $3 \times 3$ matrix, $N=3$.
2. Determinant of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{N-1}$. Again, $N$ is the order of the matrix.
3. Iterative application of the adjoint property: If we apply the adjoint property multiple times, say $k$ times, for an $N \times N$ matrix $A$: $|\operatorname{adj}(\operatorname{adj}(\dots(\operatorname{adj} A)\dots))| \text{ (k times)} = |A|^{(N-1)^k}$

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Step-by-Step Solution

Given the matrix $A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]$. The order of matrix $A$ is $N=3$.

Step 1: Calculate the Determinant of Matrix A, $|A|$

We begin by finding the determinant of the given matrix $A$. We can expand the determinant along the first row for simplicity, as it contains a zero element. $|A| = \left|\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right|$ Expanding along the first row: $|A| = 2 \cdot \left|\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right| - 1 \cdot \left|\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right| + 0 \cdot \left|\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right|$ Now, we calculate the $2 \times 2$ determinants: $|A| = 2 \cdot ((2)(2) - (-1)(-1)) - 1 \cdot ((1)(2) - (-1)(0)) + 0$ $|A| = 2 \cdot (4 - 1) - 1 \cdot (2 - 0)$ $|A| = 2 \cdot (3) - 1 \cdot (2)$ $|A| = 6 - 2$ $|A| = 4$

Step 2: Calculate the Determinant of $2A$, $|2A|$

Next, we need to find the determinant of the matrix $2A$. We use the property $|kA| = k^N |A|$. In this case, $k=2$ and $N=3$ (since $A$ is a $3 \times 3$ matrix). $|2A| = 2^3 |A|$ Substitute the value of $|A|=4$ that we calculated in Step 1: $|2A| = 8 \times 4$ $|2A| = 32$

Step 3: Simplify the Expression $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|$

Let $X = 2A$. We need to evaluate $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} X))|$. Since $A$ is a $3 \times 3$ matrix, $2A$ (which is $X$) is also a $3 \times 3$ matrix. So, the order of matrix $X$ is $N=3$. We apply the iterative adjoint property for $k=3$ adjoints: $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} X))| = |X|^{(N-1)^3}$ Substitute $N=3$ into the formula: $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} X))| = |X|^{(3-1)^3}$ $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} X))| = |X|^{2^3}$ $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} X))| = |X|^8$ Now, substitute back $X = 2A$: $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2A))| = |2A|^8$

Step 4: Substitute Values and Solve for $n$

From Step 2, we found that $|2A| = 32$. Substitute this value into the simplified expression from Step 3: $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2A))| = (32)^8$ The problem statement gives us the equation $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))| = (16)^n$. Therefore, we equate the two expressions: $(16)^n = (32)^8$ To solve for $n$, we need to express both sides of the equation with a common base. The simplest common base for 16 and 32 is 2. We know that $16 = 2^4$ and $32 = 2^5$. Substitute these into the equation: $(2^4)^n = (2^5)^8$ Using the exponent rule $(a^b)^c = a^{bc}$: $2^{4n} = 2^{5 \times 8}$ $2^{4n} = 2^{40}$ Since the bases are equal, the exponents must also be equal: $4n = 40$ Divide by 4 to solve for $n$: $n = \frac{40}{4}$ $n = 10$

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Common Mistakes & Tips

• Order of Matrix (N): Always correctly identify the order of the matrix ($N$) at each step. For this problem, $A$ is $3 \times 3$, so $N=3$. This is fundamental for applying the determinant properties correctly.
• Exponent Rules: Be meticulous with exponent rules. The iterative adjoint property involves powers of $(N-1)$, specifically $(N-1)^k$ for $k$ adjoints, not $k(N-1)$.
• Base Conversion: When solving exponential equations like $a^x = b^y$, convert both $a$ and $b$ to their smallest common prime factor base (e.g., $16=2^4$, $32=2^5$) to simplify calculations and prevent errors.

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Summary

This problem is a direct test of your understanding and application of two fundamental properties of determinants: the determinant of a scalar multiple of a matrix ($|kA| = k^N |A|$) and the determinant of the adjoint of a matrix ($|\operatorname{adj} A| = |A|^{N-1}$). By systematically calculating $|A|$, then $|2A|$, and finally applying the iterative adjoint property for three adjoints, we simplified the given expression. Equating this to $(16)^n$ and converting both sides to a common base (base 2) allowed us to solve for $n$ using basic exponent rules.

The final answer is $\boxed{10}$, which corresponds to option (C).