How to quickly compute the determinant of given matrix I need to find the determinant of given 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 know that It can be computed with the help of row operations;by applying
R1 $\to$ R1 + R2+R3+R4+R5+R6, and then subsequently simplifying using further row operations to find the determinant.
However my question is : 
Is there another way to compute the determinant quickly ?,because the row operations are quite time taking and error prone.
I thought of using eigenvalues here but could not reach the solution
Can anyone help me to find this determinant quickly ?
Thank you

By reordering both rows and columns you can see that the determinant is equal to that of a block diagonal matrix:
$$
\begin{vmatrix}
1&2&0&0&0&0\\
2&1&0&0&0&0\\
0&0&1&2&0&0\\
0&0&2&1&0&0\\
0&0&0&0&1&2\\
0&0&0&0&2&1\\
\end{vmatrix}
= \left(\begin{vmatrix}
1&2 \\ 2&1
\end{vmatrix}\right)^3
$$

Or with Schur complements: We have
$$
 M = \begin{pmatrix}
A & B \\ C &D
\end{pmatrix}  \text{ with } A = D = \begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 1
\end{pmatrix}  \text{ and } B = C = \begin{pmatrix}
0 & 0 & 2 \\ 0 & 2 & 0 \\ 2& 0 & 0
\end{pmatrix}
$$
and therefore (with $I_3$ denoting the $3\times 3$ identity matrix):
$$
 \det M = \det(A) \det(D - CA^{-1}B) = \det(I_3 - B^2) = \det(-3I_3) = -27 \, .
$$
