<s>[INST] Given a matrix $B$, what is $det(B^4)$? My task is this:
Compute det$(B^4)$, where 
$$B =\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix}$$
My work so far:
It can be shown that 
$B = 
\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\end{pmatrix}\implies B$ is invertable.
Applying theorem for determinants to triangular matrices we get det$(B) = 1*1*(-4)=-4$.
Appling the multiplicative property theorem, we should get that det$(B^4)=$det$(B)$det$(B)$det$(B)$det$(B)=[$det$(B)]^4=(-4)^4= 256.$
Now I need help from the community to verify if the reasoning is correct, because I've already had a fellow student telling me that the answer is 16, and indeed the textbook I'm using states the same, which makes me abit confused. So please point out my errors or verify that each step holds true.
Thanks in advance.
 [/INST] I don't know how you have obtained 
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\end{pmatrix}$$
But making elementary operations ($R2\leftarrow R2-R1$, $R3\leftarrow R3-R1$) I obtain
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 1\\
0 & 2 & 0
\end{pmatrix}$$
and then ($R3\leftarrow R3-2\cdot R2$):
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 1\\
0 & 0 & -2
\end{pmatrix}$$
</s>