Answer
\[
\operatorname{det}\left(B^{4}\right)=(-2)^{4}=16
\]
Work Step by Step
$B=\left(\begin{array}{lll}1 & 0 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 1\end{array}\right) \sim\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -2\end{array}\right) \rightarrow \operatorname{det}(B)=-2$
$R 3-R 1 \rightarrow R 2-R 1 \rightarrow R 3-2 R 2 . \quad B$ is now in
triangular form so apply theorem 2 from 3.1 .
\[
\begin{aligned}
\operatorname{det}\left(B^{4}\right) &=\operatorname{det}(B B B B) \\
&=\operatorname{det}(B) \operatorname{det}(B) \operatorname{det}(B) \operatorname{det}(B) \\
&=[\operatorname{det}(B)]^{4}
\end{aligned}
\]
Apply multiplicative property theorem (6) from 3.2 to get the result.
