Question

Let \[ X= \begin{pmatrix} 1 & a & 3 \\ 1 & 3 & 3 \\ 4 & 2 & 2 \end{pmatrix} \] be the adjoint of a \(3 \times 3\) matrix \(A\), and let \[ \det(A)=6. \] Then, the value of \(a\) is:

• A. 33/10
• B. 0
• C. 66/5
• D. 33/5

Hint:

Remember the general rule: $|\text{adj}(A)| = |A|^{n-1}$. For a $3\times3$ matrix, the determinant of the adjoint is always the square of the determinant of the original matrix.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For any $n \times n$ matrix $A$, the determinant of its adjoint matrix, $\text{adj}(A)$, is related to the determinant of $A$ by a fundamental algebraic theorem.
Key Formula or Approach:
The relation is given by:
\[ |\text{adj}(A)| = |A|^{n-1} \]

Step 2: Detailed Explanation:
We are given:
$n = 3$ (since $A$ is a $3 \times 3$ matrix)
$|A| = 6$
$X = \text{adj}(A)$
Using the formula, compute the determinant of $X$:
\[ |X| = |\text{adj}(A)| = |A|^{3-1} = |A|^2 = 6^2 = 36 \]
Now, let us calculate the determinant of $X$ directly from the given matrix:
\[ |X| = \begin{vmatrix} 1 & a & 3 1 & 3 & 3 4 & 2 & 2 \end{vmatrix} \]
Expand along the first row:
\[ |X| = 1 \cdot (3 \cdot 2 - 3 \cdot 2) - a \cdot (1 \cdot 2 - 3 \cdot 4) + 3 \cdot (1 \cdot 2 - 3 \cdot 4) \]
\[ |X| = 1 \cdot (6 - 6) - a \cdot (2 - 12) + 3 \cdot (2 - 12) \]
\[ |X| = 0 - a \cdot (-10) + 3 \cdot (-10) = 10a - 30 \]
Equate the calculated determinant to the theoretical value 36:
\[ 10a - 30 = 36 \]
\[ 10a = 66 \implies a = \frac{66}{10} = \frac{33}{5} \]
Therefore, the value of $a$ is $33/5$.

Step 3: Final Answer
The correct option is (D).