MCQ
Let $A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]$. If $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^{ n }$, then $n$ is equal to
- ✓$10$
- B$9$
- C$12$
- D$8$
$\therefore|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A ))|$
$=|2 A |^{( n -1)^3} \Rightarrow|2 A |^8=16^n$
$\Rightarrow\left(2^3|A|\right)^8=16^n$
$\Rightarrow\left(2^3 \times 2^2\right)^8=16^n$
$=2^{40}=16^n$
$=16^{10}=16^n \Rightarrow n=10$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f (\theta)=\left|\begin{array}{ccc}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{array}\right|$ are $m$ and $M$ respectively, then the ordered pair $( m , M )$ is equal to