Key Concepts and Formulas for Matrices and Determinants

This problem requires a strong understanding of properties related to determinants and adjoints of matrices. For an $m \times m$ matrix $A$, the following properties are crucial:

1. Determinant of a scalar multiple: $|kA| = k^m |A|$
2. Determinant of the adjoint: $|\text{adj}(A)| = |A|^{m-1}$
3. Adjoint of a scalar multiple: $\text{adj}(kA) = k^{m-1} \text{adj}(A)$
4. Adjoint of the adjoint: $\text{adj}(\text{adj}(A)) = |A|^{m-2} A$
5. Determinant of the adjoint of the adjoint: This can be derived from property 4: $|\text{adj}(\text{adj}(A))| = ||A|^{m-2} A|$ Using property 1, where the scalar is $|A|^{m-2}$: $|\text{adj}(\text{adj}(A))| = (|A|^{m-2})^m |A|$ $|\text{adj}(\text{adj}(A))| = |A|^{m(m-2)} |A|$ $|\text{adj}(\text{adj}(A))| = |A|^{m^2-2m+1} = |A|^{(m-1)^2}$

In this specific problem, the matrix $A$ is a $3 \times 3$ matrix, so $m=3$. Let's adapt the general formulas for $m=3$:

1. $|kA| = k^3 |A|$
2. $|\text{adj}(A)| = |A|^{3-1} = |A|^2$
3. $\text{adj}(kA) = k^{3-1} \text{adj}(A) = k^2 \text{adj}(A)$
4. $\text{adj}(\text{adj}(A)) = |A|^{3-2} A = |A| A$
5. $|\text{adj}(\text{adj}(A))| = |A|^{(3-1)^2} = |A|^4$

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Step 1: Determine the value of $\alpha$

We are given the matrix $A$ and its determinant $|A|=2$. Our first goal is to use this information to find the value of $\alpha$.

Given matrix: $A = \begin{bmatrix} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$

We calculate the determinant of $A$ using cofactor expansion along the first row. $|A| = 1 \cdot \left| \begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix} \right| - 2 \cdot \left| \begin{matrix} \alpha & 1 \\ 1 & 2 \end{matrix} \right| + 3 \cdot \left| \begin{matrix} \alpha & 3 \\ 1 & 1 \end{matrix} \right|$ Let's calculate each term:

• The first term is $1 \times (\text{determinant of minor for } A_{11})$: $1 \times (3 \times 2 - 1 \times 1) = 1(6 - 1) = 5$
• The second term is $-2 \times (\text{determinant of minor for } A_{12})$: $-2 \times (\alpha \times 2 - 1 \times 1) = -2(2\alpha - 1) = -4\alpha + 2$
• The third term is $+3 \times (\text{determinant of minor for } A_{13})$: $+3 \times (\alpha \times 1 - 3 \times 1) = +3(\alpha - 3) = 3\alpha - 9$ Now, sum these terms to find $|A|$: $|A| = 5 + (-4\alpha + 2) + (3\alpha - 9)$ $|A| = 5 - 4\alpha + 2 + 3\alpha - 9$ $|A| = (5 + 2 - 9) + (-4\alpha + 3\alpha)$ $|A| = -2 - \alpha$ We are given that $|A|=2$. So, we can set up the equation: $-2 - \alpha = 2$ $-\alpha = 2 + 2$ $-\alpha = 4$ $\alpha = -4$ Thus, the value of $\alpha$ is $-4$.

Tip: When calculating determinants, always be careful with signs in cofactor expansion. For a $3 \times 3$ matrix, the sign pattern is $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$.

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Step 2: Simplify the expression $|2 \text{adj}(2 \text{adj}(2A))|$

Our next goal is to simplify the given complex determinant expression using the properties derived above. We will work from the innermost part outwards.

Let the expression be $E = |2 \text{adj}(2 \text{adj}(2A))|$. We know that for $m=3$, $|A|=2$.

1. Simplify the innermost term, $\text{adj}(2A)$: Using the property $\text{adj}(kA) = k^2 \text{adj}(A)$ (for $m=3$): $\text{adj}(2A) = 2^2 \text{adj}(A) = 4 \text{adj}(A)$

2. Simplify the next layer, $2 \text{adj}(2A)$: Substitute the result from step 1: $2 \text{adj}(2A) = 2 (4 \text{adj}(A)) = 8 \text{adj}(A)$

3. Simplify $\text{adj}(2 \text{adj}(2A))$: This is equivalent to $\text{adj}(8 \text{adj}(A))$. Again, we use the property $\text{adj}(kM) = k^2 \text{adj}(M)$, where $M = \text{adj}(A)$ and $k=8$: $\text{adj}(8 \text{adj}(A)) = 8^2 \text{adj}(\text{adj}(A))$ $= 64 \text{adj}(\text{adj}(A))$ Now, use the property $\text{adj}(\text{adj}(A)) = |A|A$ (for $m=3$): $= 64 (|A|A)$ Since $|A|=2$: $= 64 (2A) = 128A$

4. Simplify $2 \text{adj}(2 \text{adj}(2A))$: Substitute the result from step 3: $2 \text{adj}(2 \text{adj}(2A)) = 2 (128A) = 256A$

5. Calculate the determinant $|2 \text{adj}(2 \text{adj}(2A))|$: This is equivalent to $|256A|$. Using the property $|kA| = k^3 |A|$ (for $m=3$), where $k=256$: $|256A| = (256)^3 |A|$ Substitute the value $|A|=2$: $= (256)^3 \cdot 2$ To compare this with $32^n$, let's express it in powers of 2. We know $256 = 2^8$: $= (2^8)^3 \cdot 2^1$ $= 2^{24} \cdot 2^1$ $= 2^{25}$ So, we have found that $|2 \text{adj}(2 \text{adj}(2A))| = 2^{25}$.

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Step 3: Equating to $32^n$ and finding $n$

We are given that $|2 \text{adj}(2 \text{adj}(2A))| = 32^n$. From Step 2, we found that $|2 \text{adj}(2 \text{adj}(2A))| = 2^{25}$. Equating these two expressions: $2^{25} = 32^n$ We need to express $32$ as a power of $2$. We know $32 = 2^5$: $2^{25} = (2^5)^n$ Using the exponent rule $(a^b)^c = a^{bc}$: $2^{25} = 2^{5n}$ Since the bases are equal, the exponents must be equal: $25 = 5n$ $n = \frac{25}{5}$ $n = 5$ Thus, the value of $n$ is $5$.

Common Mistake: A common error is misapplying the determinant properties, especially confusing $k^m$ with $k^{m-1}$ or forgetting the scalar multiple inside the adjoint. Always write down the properties for the specific matrix order ($m=3$ here) before starting.

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Step 4: Calculate $3n + \alpha$

Finally, we need to calculate the value of $3n + \alpha$. From Step 1, we found $\alpha = -4$. From Step 3, we found $n = 5$. Substitute these values into the expression: $3n + \alpha = 3(5) + (-4)$ $= 15 - 4$ $= 11$

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Summary and Key Takeaway

The final value of $3n + \alpha$ is $11$. This problem is an excellent test of your mastery of matrix properties, particularly those involving determinants and adjoints. The key to solving such problems is to:

1. Systematically apply the properties by working from the innermost expressions outwards.
2. Keep track of the matrix order ($m=3$ in this case), as it dictates the exponents in the formulas.
3. Perform careful algebraic manipulations and conversions to common bases (like base 2 in this problem) when dealing with powers.