Question

If $A$ is a $3\times 3$ matrix such that $|A|=27,\text{Adj}A=k{A}^T$, then $k^2-3k+5=$

• A. 5
• B. 3
• C. 0
• D. 2

Answer 1

Answer: (A)

$|A|=27$
$\mathrm{adj}A=k{A}^T$
$\Rightarrow |\mathrm{adj}A|=k^3\left|A^T\right|$
$\Rightarrow |A{|}^2=k^3|A|$
$\Rightarrow |A|\left(|A|-k^3\right)=0$
Since, $|A|\ne 0$
$\Rightarrow k^3=|A|=27$
$k=3$
$k^2-3k+5=9-9+5=5$