Incorrectly solving the determinant of a matrix

Question

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

Answer 1

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 . $

Answer 2

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 $$ \det(B^4) = \det(B)^4 = (-2)^4 = 16 $$

Answer 3

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 $\;\det B^4=(\det B)^4=16$.

Answer 4

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 $(-2)^4=16$