Mathematics - Matrix Question with Solution | TestHub

Given $A=\left[\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ 

Now $A^2=\left[\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$

i.e. $A^{98}=I;A^{49}=A$

$\therefore B_0=A^{98}+2A^{49}=I+2A=\left[\begin{array}{lll}1& 2& 0\\ 2& 1& 0\\ 0& 0& 3\end{array}\right]$

Given $B_n=adj\left(B_{n-1}\right)$

$\therefore B_4=adj\left(B_3\right)=adj\left(adj{B}_2\right)=adj\left(adj\left(adj{B}_1\right)\right)$

$=adj\left(adj\left(adj\left(adj{B}_0\right)\right)\right)$

$\therefore \left|B_4\right|={\left|B_0\right|}^{2^4}={\left|B_0\right|}^{16}$

Now $\left|B_0\right|=\left[\begin{array}{lll}1& 2& 0\\ 2& 1& 0\\ 0& 0& 3\end{array}\right]=-9$

Hence, $\left|B_4\right|={\left|B_0\right|}^{16}=9^{16}=3^{32}$