Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B = 
    \begin{bmatrix}
     1       & 0 & 1 \\
     1       & 1 & 2 \\
     1       & 2 & 1 
    \end{bmatrix}
$ 
I created
$C=\begin{bmatrix}
     1       & 2 \\
     1       & 1 \\
    \end{bmatrix}
$ and 
$
 D= \begin{bmatrix}
     1       & 1 \\
     1       & 2 \\
    \end{bmatrix}
$
$det(B) = -det(C) -2det(D) = -(1-2) - 2(2-1) = -(-1)-2(1)=1-2=1, 1^4=1$
However, the correct answer is 16. I'm confused on where I made my wrong turn 

Why did you do that? Just use determinant along the first row. This way is easier because there is a 0 in that row.
$$
        \begin{vmatrix}
        1 & 0 & 1 \\
        1 & 1 & 2 \\
        1 & 2 & 1 \\
        \end{vmatrix}=1\begin{vmatrix}
        1 & 2 \\
        2 & 1 \\
        \end{vmatrix}-0\begin{vmatrix}
        1 & 2 \\
        1 & 1 \\
        \end{vmatrix}+1\begin{vmatrix}
        1 & 1 \\
        1 & 2 \\
        \end{vmatrix}=(1*1-2*2)+(1*2-1*1)=\\
-3+1=-2
$$
and $(-2)^4=16$
