1. Fundamental Properties of Determinants and Adjoints

This problem relies on a few key properties of determinants and adjoints for an $n \times n$ matrix $X$:

1. Determinant of an Adjoint Matrix: The determinant of the adjoint of a matrix $X$ is given by: $|\text{adj}(X)| = |X|^{n-1}$ where $n$ is the order of the matrix.

2. Determinant of a Scalar Multiple of a Matrix: If $K$ is a scalar and $X$ is an $n \times n$ matrix, then the determinant of $KX$ is: $|KX| = K^n |X|$ where $n$ is the order of the matrix.

3. Properties of Exponents:

   • $(a^b)^c = a^{bc}$
   • $a^b \cdot a^c = a^{b+c}$
   • $\frac{a^b}{a^c} = a^{b-c}$
   • $(ab)^c = a^c b^c$

In this problem, we are given that $A$ is a $3 \times 3$ matrix, so the order $n=3$.

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2. Step-by-Step Solution

We are given the equation: $|\text{adj} (24A)| = |\text{adj} (3 \text{ adj} (2A))|$ Our goal is to find the value of $|A|^2$.

Step 1: Apply the Determinant of Adjoint Property to the Outermost adj

• Why this step? The given equation has adj operations on both sides. Applying the property $|\text{adj}(X)| = |X|^{n-1}$ (with $n=3$) will simplify the expressions by removing these outermost adj terms.

Let's apply this property to both sides:

• For the left side, let $X = 24A$. Then, $|\text{adj}(24A)| = |24A|^{3-1} = |24A|^2$.
• For the right side, let $X = 3 \text{ adj}(2A)$. Then, $|\text{adj}(3 \text{ adj}(2A))| = |3 \text{ adj}(2A)|^{3-1} = |3 \text{ adj}(2A)|^2$.

Substituting these back into the original equation, we get: $|24A|^2 = |3 \text{ adj}(2A)|^2$

Step 2: Apply the Determinant of a Scalar Multiple Property

• Why this step? We now have terms like $|KX|^2$. We need to extract the scalar multiples ($24$ and $3$) out of the determinant using the property $|KX| = K^n |X|$ (with $n=3$).

Let's apply this property:

• For the left side, $K=24$ and $X=A$. So, $|24A| = 24^3 |A|$. Therefore, $|24A|^2 = (24^3 |A|)^2 = (24^3)^2 |A|^2 = 24^{3 \times 2} |A|^2 = 24^6 |A|^2$.
• For the right side, $K=3$ and $X=\text{adj}(2A)$. So, $|3 \text{ adj}(2A)| = 3^3 |\text{adj}(2A)|$. Therefore, $|3 \text{ adj}(2A)|^2 = (3^3 |\text{adj}(2A)|)^2 = (3^3)^2 |\text{adj}(2A)|^2 = 3^{3 \times 2} |\text{adj}(2A)|^2 = 3^6 |\text{adj}(2A)|^2$.

Substituting these results back into the equation from Step 1: $24^6 |A|^2 = 3^6 |\text{adj}(2A)|^2$

Step 3: Eliminate the Remaining Adjoint Term

• Why this step? The right side still contains an adj term, $|\text{adj}(2A)|$. We need to simplify this using the determinant of adjoint property again.

Applying $|\text{adj}(X)| = |X|^{n-1}$ for $X=2A$ and $n=3$: $|\text{adj}(2A)| = |2A|^{3-1} = |2A|^2$

Substitute this into our current equation: $24^6 |A|^2 = 3^6 (|2A|^2)^2$ $24^6 |A|^2 = 3^6 |2A|^4$

Step 4: Expand the Final Determinant Term

• Why this step? The equation now involves $|2A|$. We need to expand this using the determinant of a scalar multiple property to express everything in terms of $|A|$ and constants.

Applying $|KX| = K^n |X|$ for $K=2$ and $X=A$: $|2A| = 2^n |A| = 2^3 |A|$

Substitute this into the equation from Step 3: $24^6 |A|^2 = 3^6 (2^3 |A|)^4$ $24^6 |A|^2 = 3^6 (2^{3 \times 4} |A|^4)$ $24^6 |A|^2 = 3^6 \cdot 2^{12} \cdot |A|^4$

Step 5: Solve for $|A|^2$

• Why this step? We have an algebraic equation with $|A|$ as the variable. We need to simplify the constants and isolate $|A|^2$.

First, let's simplify $24^6$ using prime factorization: $24 = 3 \times 8 = 3 \times 2^3$. So, $24^6 = (3 \times 2^3)^6 = 3^6 \times (2^3)^6 = 3^6 \times 2^{18}$.

Substitute this back into the equation: $3^6 \cdot 2^{18} \cdot |A|^2 = 3^6 \cdot 2^{12} \cdot |A|^4$

Since $A$ is an invertible matrix, its determinant $|A|$ is non-zero ($|A| \neq 0$). This allows us to divide both sides by $|A|^2$. We can also divide by $3^6$ as it's a non-zero constant.

Divide both sides by $3^6$: $2^{18} \cdot |A|^2 = 2^{12} \cdot |A|^4$

Divide both sides by $|A|^2$ (since $|A| \neq 0$): $2^{18} = 2^{12} \cdot |A|^2$

Now, isolate $|A|^2$: $|A|^2 = \frac{2^{18}}{2^{12}}$ Using the exponent property $\frac{a^b}{a^c} = a^{b-c}$: $|A|^2 = 2^{18-12}$ $|A|^2 = 2^6$

Step 6: Final Calculation

• Why this step? To provide the numerical value of $2^6$.

Calculate $2^6$: $2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

Thus, $|A|^2 = 64$. The question asks for $|A|^2$, which is $2^6$.

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3. Important Tips and Common Pitfalls

• Mind the Order of the Matrix ($n$): Always identify the order of the matrix ($n$) at the beginning. In this problem, $n=3$. Incorrectly using $n=2$ or $n=4$ would lead to wrong exponents.
• Correct Application of Scalar Property: Remember that $|KX| = K^n |X|$, not just $K|X|$. And when raising this to a power, like $(K^n |X|)^m = K^{nm} |X|^m$, ensure all terms inside the parenthesis are raised to the power. A common mistake is to write $K^n |X|^m$ instead of $K^{nm} |X|^m$.
• Exponent Rules are Crucial: Be careful with exponent rules, especially $(a^b)^c = a^{bc}$ and $a^b \cdot a^c = a^{b+c}$.
• Prime Factorization for Simplification: Breaking down constants like $24$ into prime factors ($3 \times 2^3$) greatly simplifies the algebraic manipulation and cancellation of terms.
• Invertibility Condition: The fact that $A$ is an invertible matrix means $|A| \neq 0$. This is crucial because it allows us to divide both sides of the equation by $|A|^2$ without losing solutions.

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4. Summary and Key Takeaway

This problem is a comprehensive test of your understanding of fundamental determinant properties, particularly those involving the adjoint of a matrix and scalar multiples. The solution involves systematically applying these properties from the outermost operations inwards, carefully managing exponents, and then using algebraic simplification to solve for the required term. Mastering these properties and careful execution are key to solving such problems efficiently.

The final answer is $\boxed{2^6}$.