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  4. Let the matrix A=[0 1 0 0 0 1 1 0 0] and the matrix B0=A49+2 A98. If Bn= textAdj(Bn-1) for all n ≥ 1, then textdet( B 4) is equal to :

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Q. Let the matrix $A=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}$ and the matrix $B_0=A^{49}+2 A^{98}$. If $B_n=\text{Adj}\left(B_{n-1}\right)$ for all $n \geq 1$, then $\text{det}\left( B _4\right)$ is equal to :

JEE MainJEE Main 2022Determinants

Solution:

$A^2=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}$
$=\begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$
$ a \leftrightarrow R _2$
$-\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}$
$R _2 \leftrightarrow R _3$
$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=I$
$B _0= A ^{49}+2 A ^{98} $
$ = A +2 I $
$ B _{ n } =\text{Adj}\left( B _{ n }-1\right) $
$B_4=\text{Adj}\left(\text{Adj}\left(\text{Adj}\left(\text{Adj}_0\right)\right)\right.$
$ =\left| B _0\right|^{( n -1)^4}$
$=\left| B _0\right|^{16}$
$B_0=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}+\begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$
$=\begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2\end{bmatrix}$
$ =2(4-0)-1(0-1) $
$ =9 $
$ B _4(9)^{16}=(3)^{32}$