Question

If $P = \begin{bmatrix}1&\alpha&3\\ 1&3&3\\ 2&4&4\end{bmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $| A | = 4,$ then $\alpha$ is equal to

• A. 11
• B. 5
• C. 0
• D. 4

Answer 1

Answer: (A)

$P = \begin{bmatrix}1&\alpha&3\\ 1&3&3\\ 2&4&4\end{bmatrix}$
Let $|P| = 1(12-12) - \alpha (4 - 6) + 3 ( 4 - 6 ) = 2 \alpha - 6 $
Also, det(adj A) = (det A)$^2$.
$\Rightarrow \ 2\alpha - 6 = 16 \ \ \ \Rightarrow \ 2\alpha =22.\ \ \ \ \therefore \ \ \alpha = 11$
Remark: det(adj $A$):= (det $A$)$^{n - 1}$· where A is a matrix of order n.