Q78 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics
Given $P = \begin{bmatrix} 2 & \alpha & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 3 \end{bmatrix}$ is the adjoint of a $3 \times 3$ matrix A and $|A| = 3$, then the value of $\alpha$ is:
- Correct answerA$-8$
- B$-26$
- C$7$
- D$-25$
Solution
Given $P = \text{adj}(A)$ and $A$ is a $3 \times 3$ matrix.
We know that $|\text{adj}(A)| = |A|^{n-1}$, where $n$ is the order of the matrix.
Here, $n = 3$ and $|A| = 3$.
$|P| = |A|^{3-1} = |A|^2 = 3^2 = 9$
Now, calculating the determinant of matrix $P$:
$|P| = \begin{vmatrix} 2 & \alpha & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 3 \end{vmatrix}$
Expanding along the first row:
$|P| = 2(2 \times 3 - 2 \times 3) - \alpha(1 \times 3 - 2 \times 1) + 1(1 \times 3 - 2 \times 1)$
$|P| = 2(0) - \alpha(1) + 1(1)$
$|P| = 1 - \alpha$
Equating the two values of $|P|$:
$1 - \alpha = 9$
$\alpha = -8$
Answer: $-8$
We know that $|\text{adj}(A)| = |A|^{n-1}$, where $n$ is the order of the matrix.
Here, $n = 3$ and $|A| = 3$.
$|P| = |A|^{3-1} = |A|^2 = 3^2 = 9$
Now, calculating the determinant of matrix $P$:
$|P| = \begin{vmatrix} 2 & \alpha & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 3 \end{vmatrix}$
Expanding along the first row:
$|P| = 2(2 \times 3 - 2 \times 3) - \alpha(1 \times 3 - 2 \times 1) + 1(1 \times 3 - 2 \times 1)$
$|P| = 2(0) - \alpha(1) + 1(1)$
$|P| = 1 - \alpha$
Equating the two values of $|P|$:
$1 - \alpha = 9$
$\alpha = -8$
Answer: $-8$
Hence, the correct answer is option: 'A'.