Let [A=[[arrayccc 2 & 1 & 0 1 & 2 & -1 0 & -1 & 2 array]]].

Question

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 :

Accepted Answer

We have,

$$
\begin{aligned}
& |\mathrm{A}|=\left|\[\begin{array}{ccc} 2 & 1 & 0 \ 1 & 2 & -1 \ 0 & -1 & 2 \end{array}\end{aligned}$$\right|=2(4-1)-1(2-0)+0 \\
\& =6-2=4 \\
\& \text { So, }|2 \mathrm{~A}|=2\^3|\mathrm{~A}|=8 \times 4=32 \\
\& \text { Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|\^{(n-1)\^3} \\
\& =(32)\^{2\^3}=32\^8 \\
\& \Rightarrow 16\^n=(32)\^8=2\^8 \times 16\^8 \\
\& \Rightarrow 16\^n=16\^{2+8} \Rightarrow n=10

\]

Concepts :

(a) $|k \mathrm{~A}|=k^n|\mathrm{~A}|$

(b) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}$

Solutions · 1