Q91 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { If } P=\left[\begin{array}{lll}
1 & \alpha & 3 \\
1 & 3 & 3 \\
2 & 4 & 4
\end{array}\right] \text { is the adjoint of a } 3 \times 3 \text { matrix } A \text { and }|A|=4 \text { then } \alpha \text { is equal to }$
- A4
- B0
- C5
- Correct answerD11
Solution
The property of the adjoint of a matrix states that $|adj(A)| = |A|^{n-1}$, where $n$ is the order of the matrix $A$.
Given $n = 3$ and $|A| = 4$, we have $|adj(A)| = |A|^{3-1} = |A|^2 = 4^2 = 16$.
The matrix $P$ is given as $P = adj(A) = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$.
Calculating the determinant of $P$:
$|P| = 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$|P| = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$|P| = 0 - \alpha(-2) + 3(-2)$
$|P| = 2\alpha - 6$.
Equating $|P|$ to $16$:
$2\alpha - 6 = 16$
$2\alpha = 22$
$\alpha = 11$.
Answer: 11
Given $n = 3$ and $|A| = 4$, we have $|adj(A)| = |A|^{3-1} = |A|^2 = 4^2 = 16$.
The matrix $P$ is given as $P = adj(A) = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$.
Calculating the determinant of $P$:
$|P| = 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$|P| = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$|P| = 0 - \alpha(-2) + 3(-2)$
$|P| = 2\alpha - 6$.
Equating $|P|$ to $16$:
$2\alpha - 6 = 16$
$2\alpha = 22$
$\alpha = 11$.
Answer: 11
Hence, the correct answer is option: 'D'.