Unable to compute the determinant of a matrix

Question

Find the determinant of the matrix

\begin{bmatrix} 1 & 0 & 0&0&0 &2\\0&1&0&0&2&0 \\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{bmatrix}

I tried using Laplace Expansion

I got

\begin{bmatrix} 1& 0\\0&1\end{bmatrix}\begin{bmatrix} 1 &2&0 & 0\\2&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}-\begin{bmatrix} 1& 0\\0&2\end{bmatrix}\begin{bmatrix} 0 &1&2 & 0\\0&2&1&0\\2&0&0&0\\0&0&0&1\end{bmatrix}-\begin{bmatrix} 0& 2\\1&0\end{bmatrix}\begin{bmatrix} 0 &1&2 & 0\\0&2&1&0\\0&0&0&1\\2&0&0&0\end{bmatrix}+\begin{bmatrix} 0& 2\\2&0\end{bmatrix}\begin{bmatrix} 0 &0&1 & 2\\0&0&2&1\\0&2&0&0\\2&0&0&0\end{bmatrix}

The result I am getting is $-3+12-12+48=45

which is wrong

The correct answer is $-27$.

Please help where i am wrong.

NOTE:

Please dont give another method for solving this problems.I know there are many.

How to fix this problem

EDIT:

$(-1)^{1+1+2+2}$ \begin{bmatrix} 1& 0\\0&1\end{bmatrix}\begin{bmatrix} 1 &2&0 & 0\\2&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} $+(-1)^{1+2+1+3}$\begin{bmatrix} 1& 0\\0&0\end{bmatrix}\begin{bmatrix} 0 &0&2 & 0\\2&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}

+$(-1)^{1+2+1+4}$\begin{bmatrix} 1& 0\\0&0\end{bmatrix}\begin{bmatrix} 0 &0&2 & 0\\1&2&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}+...

Answer 1

Let $$ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad A = \begin{bmatrix} 0 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 0 \end{bmatrix} $$ Then your matrix can be written as $$ M = \begin{bmatrix} I & A \\ A & I \end{bmatrix}. $$ Then a formular gives us $$ \det M = \det I \cdot \det(I - A^2). $$ Since $$ I - A^2 = I - 4I = -3I, $$ we conclude that $$ \det M = -3^3 = -27.$$

Answer 2

We denote the given matrix by $A$ and the field over which we are working by $K$.

If $\operatorname{char}(K) = 2$ then $A$ is the identity matrix, and it follows that $\det(A) = 1$.

We now consider the case $\operatorname{char}(K) \neq 2$: Note that the matrix $$ A + I = \begin{bmatrix} 2 & 0 & 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & 0 & 2 & 0 \\ 0 & 0 & 2 & 2 & 0 & 0 \\ 0 & 0 & 2 & 2 & 0 & 0 \\ 0 & 2 & 0 & 0 & 2 & 0 \\ 2 & 0 & 0 & 0 & 0 & 2 \end{bmatrix} $$ has rank $3$, so that $\dim \ker(A + I) = 3$. This shows that $-1$ is an eigenvalue of $A$ with multiplicity $3$. Similarly, the matrix $$ A - 3I = \begin{bmatrix} -2 & 0 & 0 & 0 & 0 & 2 \\ 0 & -2 & 0 & 0 & 2 & 0 \\ 0 & 0 & -2 & 2 & 0 & 0 \\ 0 & 0 & 2 & -2 & 0 & 0 \\ 0 & 2 & 0 & 0 & -2 & 0 \\ 2 & 0 & 0 & 0 & 0 & -2 \end{bmatrix} $$ has rank $3$, so that $\dim \ker(A - 3I) = 3$. This shows that $3$ is an eigenvalue of $A$ with multiplicity $3$.

We have that $-1 \neq 3$ because $\operatorname{char}(K) \neq 2$. It follows that $A$ is diagonalizable with eigenvalues $-1, -1, -1, 3, 3, 3$. The determianent $\det(A)$ is the product of these eigenvalues, so $$ \det(A) = (-1) \cdot (-1) \cdot (-1) \cdot 3 \cdot 3 \cdot 3 = -27. $$