Key Concepts and Formulas

For an $n \times n$ matrix $X$ and a scalar $k$:

• Determinant of a scalar multiple: $|kX| = k^n |X|$
• Determinant of an adjoint matrix: $|adj(X)| = |X|^{n-1}$
• Determinant of a product: $|XY| = |X||Y|$
• Determinant of a power: $|X^m| = |X|^m$

Given:

• $A$ is a $3 \times 3$ matrix, which means its dimension $n=3$.
• $|A| = 2$.

We need to calculate $|3\,adj\,(|3A|{A^2})|$. We will break this down step-by-step from the innermost expression outwards.

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Step 1: Evaluate the innermost determinant, $|3A|$. We use the property for the determinant of a scalar multiple, $|kX| = k^n|X|$. In this case, $X=A$ and the scalar $k=3$. Since $A$ is a $3 \times 3$ matrix, its dimension $n=3$. $|3A| = 3^n |A| = 3^3 |A|$ Substitute the given value of $|A|=2$: $|3A| = 27 \times 2 = 54$ Explanation: The first step is to simplify the scalar value $|3A|$ that is part of the argument for the adjoint function. It's crucial to remember that the scalar factor $3$ is raised to the power of the matrix dimension, $n=3$.

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Step 2: Define the matrix argument for the adjoint function. Let the matrix inside the $adj$ function be $M$. $M = |3A|A^2$ Substitute the value of $|3A|=54$ from Step 1: $M = 54 A^2$ Explanation: The term $54$ is now a scalar. When a scalar multiplies a matrix ($A^2$), the result is another matrix of the same dimension. Since $A$ is $3 \times 3$, $A^2$ is also $3 \times 3$, and therefore $M = 54A^2$ is a $3 \times 3$ matrix.

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Step 3: Apply the scalar multiplication property to the outermost determinant. The expression we need to evaluate is $|3\,adj(M)|$. Here, $adj(M)$ is a matrix. Let $Y = adj(M)$. We are evaluating $|3Y|$. Since $M$ is a $3 \times 3$ matrix, $adj(M)$ is also a $3 \times 3$ matrix. So, the dimension $n$ for $adj(M)$ is $3$. Using the property $|kY| = k^n|Y|$ with $k=3$ and $n=3$: $|3\,adj(M)| = 3^3 |adj(M)| = 27 |adj(M)|$ Explanation: The outermost determinant involves multiplying the matrix $adj(M)$ by a scalar $3$. The scalar factor $3$ is raised to the power of the matrix's dimension, which is $3$.

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Step 4: Apply the determinant of an adjoint matrix property. Now we need to evaluate $|adj(M)|$. The property for the determinant of an adjoint matrix is $|adj(X)| = |X|^{n-1}$. Here, $X=M$. Since $M$ is a $3 \times 3$ matrix, its dimension $n=3$. $|adj(M)| = |M|^{3-1} = |M|^2$ Substitute this back into the expression from Step 3: $|3\,adj(M)| = 27 |M|^2$ Explanation: We replace $|adj(M)|$ using its definition, where the power is one less than the matrix dimension.

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Step 5: Evaluate $|M|$. From Step 2, we know $M = 54 A^2$. We need to find its determinant: $|M| = |54 A^2|$ Using the property $|kX| = k^n|X|$ again, this time with scalar $k=54$ and matrix $X=A^2$. Since $A^2$ is a $3 \times 3$ matrix, its dimension $n=3$. $|M| = 54^3 |A^2|$ Next, we need to evaluate $|A^2|$. The property for the determinant of a power is $|X^m| = |X|^m$. $|A^2| = |A|^2$ Substitute the given value of $|A|=2$: $|A^2| = 2^2 = 4$ Now, substitute this value back into the expression for $|M|$: $|M| = 54^3 \times 4$ Explanation: We first extract the scalar $54$ from the determinant, raising it to the power of $3$ (the dimension of $A^2$). Then, we evaluate the determinant of $A^2$ using the power property.

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Step 6: Substitute $|M|$ back into the main expression and simplify. From Step 4, we have $|3\,adj(M)| = 27 |M|^2$. Substitute the value of $|M| = 54^3 \times 4$ from Step 5: $|3\,adj(M)| = 27 (54^3 \times 4)^2$ Now, simplify the numerical expression using exponent rules $(ab)^c = a^c b^c$ and $(a^b)^c = a^{bc}$: $|3\,adj(M)| = 27 \times (54^3)^2 \times 4^2$ $|3\,adj(M)| = 27 \times 54^6 \times 16$ To match the format of the options, we express the numbers in terms of their prime factors (2 and 3), as $54 = 2 \times 27 = 2 \times 3^3$: $27 = 3^3$ $54 = 2 \times 3^3$ $16 = 2^4$ Substitute these into the expression: $|3\,adj(M)| = 3^3 \times (2 \times 3^3)^6 \times 2^4$ Apply the exponent rules $(ab)^c = a^c b^c$ and $(a^b)^c = a^{bc}$: $|3\,adj(M)| = 3^3 \times (2^6 \times (3^3)^6) \times 2^4$ $|3\,adj(M)| = 3^3 \times 2^6 \times 3^{18} \times 2^4$ Combine the powers of $3$ and $2$ separately: $|3\,adj(M)| = 3^{3+18} \times 2^{6+4}$ $|3\,adj(M)| = 3^{21} \times 2^{10}$

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Step 7: Compare with the given options. Let's expand each option to see which one matches our calculated value $3^{21} \times 2^{10}$: (A) $3^{12} \cdot 6^{10} = 3^{12} \cdot (2 \cdot 3)^{10} = 3^{12} \cdot 2^{10} \cdot 3^{10} = 3^{12+10} \cdot 2^{10} = 3^{22} \cdot 2^{10}$ (B) $3^{11} \cdot 6^{10} = 3^{11} \cdot (2 \cdot 3)^{10} = 3^{11} \cdot 2^{10} \cdot 3^{10} = 3^{11+10} \cdot 2^{10} = 3^{21} \cdot 2^{10}$ (C) $3^{12} \cdot 6^{11} = 3^{12} \cdot (2 \cdot 3)^{11} = 3^{12} \cdot 2^{11} \cdot 3^{11} = 3^{12+11} \cdot 2^{11} = 3^{23} \cdot 2^{11}$ (D) $3^{10} \cdot 6^{11} = 3^{10} \cdot (2 \cdot 3)^{11} = 3^{10} \cdot 2^{11} \cdot 3^{11} = 3^{10+11} \cdot 2^{11} = 3^{21} \cdot 2^{11}$

Our calculated value $3^{21} \times 2^{10}$ matches option (B).

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Common Mistakes to Avoid:

• Confusing the order of the matrix ($n$): Always be mindful of the dimension ($n=3$ in this case) for each matrix involved when applying properties like $|kX|=k^n|X|$ and $|adj(X)|=|X|^{n-1}$.
• Incorrect exponent for scalar multiplication: A common error is to use $k|X|$ or $k^{n-1}|X|$ instead of the correct $k^n|X|$ for $|kX|$.
• Incorrect exponent for adjoint determinant: Another common mistake is to use $|X|^n$ instead of $|X|^{n-1}$ for $|adj(X)|$.
• Careless numerical simplification: Ensure careful application of exponent rules like $(a^b)^c = a^{bc}$ and $(ab)^c = a^c b^c$. Break down numbers into prime factors to avoid errors and easily compare with options.
• Order of operations: Always work from the innermost part of the expression outwards, carefully identifying whether a term is a scalar or a matrix at each stage.

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Summary / Key Takeaway: This problem effectively tests a student's thorough understanding and careful application of fundamental determinant properties. The solution requires a methodical approach, breaking down the complex expression into smaller, manageable parts. Mastering the properties of scalar multiplication and adjoints for determinants is crucial for solving such problems efficiently and accurately.

The final answer is $\boxed{\text{B}}$.