Q111 / 180 · COMEDK 2015 · Mathematics

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

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

Solution

Since, $\boldsymbol{A}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]$ $\Rightarrow|A|=a \cdot a \cdot a=a^{3}$ Using formula $|\operatorname{adj} A|=|A|^{n-1}$, we get $\operatorname{det}(\operatorname{adj} A)=\left(a^{3}\right)^{3-1}=\left(a^{3}\right)^{2}=a^{6}$

Hence, the correct answer is option: 'C'.

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