Key Concepts and Formulas

To solve this problem, we need to apply several fundamental properties of determinants for a square matrix $A$ of order $n$. In this specific problem, $A$ is a $3 \times 3$ matrix, so $n=3$.

1. Determinant of a scalar multiple: For any scalar $k$, $|kA| = k^n |A|$.
   • For $n=3$, this becomes $|kA| = k^3 |A|$.
2. Determinant of a product: For two square matrices $A$ and $B$ of the same order, $|AB| = |A||B|$.
3. Determinant of the adjugate matrix: For a square matrix $A$ of order $n$, $|\operatorname{adj}(A)| = |A|^{n-1}$.
   • For $n=3$, this becomes $|\operatorname{adj}(A)| = |A|^{3-1} = |A|^2$.

We are given that $|A|=5$.

---

Step-by-Step Solution

Our goal is to evaluate the expression $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|$ and express it in the form $2^\alpha \cdot 3^\beta \cdot 5^\gamma$. We will systematically simplify the expression by working from the outermost determinant inwards, applying the properties listed above.

Step 1: Apply the scalar multiplication property to the outermost determinant. The outermost expression is $|2 \cdot X|$, where $X = \operatorname{adj}(3 A \operatorname{adj}(2 A))$. Using the property $|kX| = k^n |X|$ with $k=2$ and $n=3$: $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))| = 2^3 |\operatorname{adj}(3 A \operatorname{adj}(2 A))|$

Step 2: Apply the adjugate determinant property to the remaining expression. Now we have $2^3 |\operatorname{adj}(Y)|$, where $Y = 3 A \operatorname{adj}(2 A)$. Using the property $|\operatorname{adj}(Y)| = |Y|^{n-1}$ with $n=3$: $2^3 |\operatorname{adj}(3 A \operatorname{adj}(2 A))| = 2^3 |3 A \operatorname{adj}(2 A)|^2$

Step 3: Apply the determinant of a product property inside the square. The expression inside the square is $|3 A \operatorname{adj}(2 A)|$. This is the determinant of a product of two matrices: $(3A)$ and $(\operatorname{adj}(2A))$. Using the property $|UV| = |U||V|$: $2^3 |3 A \operatorname{adj}(2 A)|^2 = 2^3 (|3A| \cdot |\operatorname{adj}(2A)|)^2$

Step 4: Evaluate the individual determinants within the expression. We need to find the values of $|3A|$ and $|\operatorname{adj}(2A)|$.

• Evaluate $|3A|$: Using the scalar multiplication property $|kA| = k^n |A|$ with $k=3$ and $n=3$: $|3A| = 3^3 |A|$

• Evaluate $|\operatorname{adj}(2A)|$: First, we find the determinant of the matrix inside the adjugate, which is $2A$. Using the scalar multiplication property $|kA| = k^n |A|$ with $k=2$ and $n=3$: $|2A| = 2^3 |A|$ Now, apply the adjugate determinant property $|\operatorname{adj}(B)| = |B|^{n-1}$ to $B=2A$ with $n=3$: $|\operatorname{adj}(2A)| = |2A|^2$ Substitute the value of $|2A|$: $|\operatorname{adj}(2A)| = (2^3 |A|)^2 = 2^{3 \times 2} |A|^2 = 2^6 |A|^2$

Step 5: Substitute these evaluated determinants back into the main expression. Substitute $|3A| = 3^3 |A|$ and $|\operatorname{adj}(2A)| = 2^6 |A|^2$ into the expression from Step 3: $2^3 (|3A| \cdot |\operatorname{adj}(2A)|)^2 = 2^3 \left( (3^3 |A|) \cdot (2^6 |A|^2) \right)^2$

Step 6: Simplify the expression inside the parentheses. Combine the terms with the same base: $2^3 \left( 3^3 \cdot 2^6 \cdot |A|^{1+2} \right)^2 = 2^3 \left( 3^3 \cdot 2^6 \cdot |A|^3 \right)^2$

Step 7: Expand the square. Apply the square exponent to each factor inside the parentheses: $2^3 \left( (3^3)^2 \cdot (2^6)^2 \cdot (|A|^3)^2 \right)$ $= 2^3 \left( 3^{3 \times 2} \cdot 2^{6 \times 2} \cdot |A|^{3 \times 2} \right)$ $= 2^3 \left( 3^6 \cdot 2^{12} \cdot |A|^6 \right)$

Step 8: Combine powers of the same base. Multiply the powers of 2: $= 2^{3+12} \cdot 3^6 \cdot |A|^6$ $= 2^{15} \cdot 3^6 \cdot |A|^6$

Step 9: Substitute the given value of $|A|$. We are given $|A|=5$. Substitute this into the expression: $= 2^{15} \cdot 3^6 \cdot 5^6$

Step 10: Identify $\alpha, \beta, \gamma$ and calculate their sum. The expression is now in the form $2^\alpha \cdot 3^\beta \cdot 5^\gamma$. By comparing, we find: $\alpha = 15$ $\beta = 6$ $\gamma = 6$

Finally, we calculate the sum $\alpha + \beta + \gamma$: $\alpha + \beta + \gamma = 15 + 6 + 6 = 27$

---

Common Mistakes & Tips

• Order of Operations: Always work from the outermost determinant or operation inwards. This systematic approach reduces errors.
• Matrix Order: Be vigilant about the order $n$ of the matrix. For a $3 \times 3$ matrix, $n=3$ must be used consistently in properties like $|kA|=k^n|A|$ and $|\operatorname{adj}(A)|=|A|^{n-1}$. A common error is to use $n-1$ where $n$ is required, or vice-versa.
• Exponent Rules: Pay close attention to exponent rules, especially when dealing with nested powers, e.g., $(a^m)^p = a^{mp}$ and $a^m \cdot a^p = a^{m+p}$.
• Careful Substitution: Ensure that each substitution is done correctly to avoid arithmetic errors that can propagate through the calculation.

---

Summary

We systematically evaluated the given determinant expression by applying the properties of determinants for scalar multiplication, matrix products, and adjugate matrices. We started from the outermost operations and worked inwards, simplifying step by step. After substituting the given value of $|A|=5$, we expressed the final result as $2^{15} \cdot 3^6 \cdot 5^6$. By comparing this with the target form $2^\alpha \cdot 3^\beta \cdot 5^\gamma$, we found $\alpha=15$, $\beta=6$, and $\gamma=6$. The sum $\alpha+\beta+\gamma$ is $15+6+6=27$.

The final answer is $\boxed{27}$, which corresponds to option (A).