Question

If $P = \begin{pmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix}$ is the adjoint of a $3 \times 3$ matrix A and $det(A) = 4$, then $\alpha =$

• A. 22
• B. 11
• C. 3
• D. 4

Hint:

$|adj(A)| = |A|^{n-1}$. This is a very common property for competitive exams.

Answer 1

The Correct Option is B

Step 1: Concept
For a matrix $A$ of order $n$, the determinant of its adjoint is given by: $|adj(A)| = |A|^{n-1}$.

Step 2: Meaning
Here $n=3$ and $|A|=4$. Therefore, $|P| = |adj(A)| = 4^{3-1} = 4^{2} = 16$.

Step 3: Analysis
Calculate $|P|$: $1(12 - 12) - \alpha(4 - 6) + 3(4 - 6) = 16$ $0 + 2\alpha - 6 = 16$ $2\alpha = 22$ $\alpha = 11$.

Step 4: Conclusion
The value of $\alpha$ that satisfies the determinant condition is 11. Final Answer: (B)