Question

If $A=\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]$, then det(adj A) is

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

Answer 1

Answer: (C)

Since $A=\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]$
$\Rightarrow |A|=a.a.a=a^3$
Using formula $[adjAl=|A{|}^{n-1}$, we get det $(adjA)=(a^3{)}^{3-1}=(a^3{)}^2=a^6$