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 
 A: You have calculated det$(B) $ wrong - you just got the order of signs mixed up, when calculating the determinant by expanding into $ 2 \times 2 $ matrices. 
It is det $(B) =  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} $. 
Hence det$(B) = 1 ( 1 - 4 ) - 0(1- 2 )  + 1 ( 2-1 )   = -3 + 1 = -2 $. 
And it seems you used the formula det$(A^n) =  ($det$(A)) ^ n $ which is correct and hence det$(B^4) = ($det$(B))^4 = ( -2 )^4 = 16 . $ 
A: 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 $(-2)^4=16$
A: We have
$$ 
\det(B) = 1 \times \det(E) + 1 \times \det(D) = (1 - 4) + (2 - 1) = -2
$$
where $ E $ is 
$$
\begin{bmatrix}
1 & 2 \\ 2 & 1 
\end{bmatrix}
$$
Then 
$$
\det(B^4) = \det(B)^4 = (-2)^4 = 16
$$
A: Use row reduction to calculate the determinant:
\begin{align}
\begin{vmatrix}1&0&1\\1&1&2\\1&2&1\end{vmatrix}=\begin{vmatrix}1&0&1\\0&1&1\\0&2&0\end{vmatrix}=\begin{vmatrix}1&0&1\\0&1&1\\0&0&-2\end{vmatrix}
\end{align}
hence $\;\det B=1\cdot1\cdot(-2)=-2$, so $\;\det B^4=(\det B)^4=16$.
