Question

If $A = \begin{bmatrix}a&0&0\\\ 0&a&0\\\ 0&0&a\end{bmatrix}$, then det(adj A) is

• A. $a^{27}$
• B. $a^{9}$
• C. $a^{6}$
• D. $a^{2}$

Answer 1

Answer: (C)

$a^{6}$

Answer 2

Since $A = \begin{bmatrix}a&0&0\\\ 0&a&0\\\ 0&0&a\end{bmatrix}$
$\Rightarrow \:\:\:\: |A| = a.a.a =a^3$
Using formula $[ {adj} \, Al = |A|^{n-1}$, we get det $( {adj} \, A)= (a^3)^{3-1} = (a^3)^2 = a^6$