How to find the determinant of this $6\times 6$ X-matrix? This question was asked in my quiz and i was unable to solve it, so I am asking it here.


Find the value of determinant of this particular matrix .
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{pmatrix}$$


I have no clue on how this kind of matrices can be solved. Can anyone give a general strategy on how to solve matrices whose size are greater that $3\times 3$?
That would be really helpful.
 A: Hint
Usually using row operations will help in reducing the determinant to something that is more manageable (like diagonal or upper triangular matrices). You should know how the row operations affect the determinant.
The row operations $-2R_1+R_6, -2R_2+R_5, -2R_3+R_4$ will give
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0&\\0&0&1&2&0&0\\0&0&0&-3&0&0\\0&0&0&0&-3&0\\0&0&0&0&0&-3\end{pmatrix}$$
Now this is an upper triangular matrix. So the determinant is...
A: Performing a Laplace expansion along the first column, we get
$$
\begin{vmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{vmatrix}
=
1\cdot
\begin{vmatrix}1&0&0&2&0\\0&1&2&0&0\\0&2&1&0&0\\2&0&0&1&0\\0&0&0&0&1\end{vmatrix}
- 2\cdot
\begin{vmatrix}0&0&0&0&2\\1&0&0&2&0\\0&1&2&0&0\\0&2&1&0&0\\2&0&0&1&0\end{vmatrix}
$$
Now both $5\times 5$ determinants can be Laplace expanded along the last column to get
$$
1\cdot 1\cdot
\begin{vmatrix}1&0&0&2\\0&1&2&0\\0&2&1&0\\2&0&0&1\end{vmatrix}
- 2\cdot 2\cdot
\begin{vmatrix}1&0&0&2\\0&1&2&0\\0&2&1&0\\2&0&0&1\end{vmatrix}
=
-3\cdot \begin{vmatrix}1&0&0&2\\0&1&2&0\\0&2&1&0\\2&0&0&1\end{vmatrix}.
$$
Now you can either repeat this procedure one more time to end up with a $2\times 2$ determinant, or notice the general pattern and prove a more general statement by induction:
Let $A_n$ be the $2n\times 2n$ matrix with ones on the main diagonal and twos on the antidiagonal. What we did above to $A_3$ works out in general as
\begin{align*}
\det(A_n) &= 1\cdot\begin{vmatrix} A_{n-1} & 0 \\ 0 & 1\end{vmatrix} 
- 2 \cdot\begin{vmatrix} 0 & 2 \\ A_{n-1} & 0\end{vmatrix} \\
&= 1\cdot 1 \cdot \det(A_{n-1}) 
- 2 \cdot 2\cdot \det(A_{n-1}) \\
&= -3\cdot\det(A_{n-1}).
\end{align*}
Applied to your matrix this yields
$$
\det(A_3) = -3\det(A_2)=(-3)^2\det(A_1) = (-3)^3 = -27
$$
and in general you get $\det(A_n)=(-3)^n$.
A: Permuting the rows and columns, we obtain a block diagonal matrix.
$$\det \begin{bmatrix} \color{red}{1} & 0 & 0 & 0 & 0 & \color{red}{2}\\ 0 & \color{orange}{1} & 0 & 0 & \color{orange}{2} & 0\\ 0 & 0 & \color{magenta}{1} & \color{magenta}{2} & 0 & 0\\ 0 & 0 & \color{magenta}{2} & \color{magenta}{1} & 0 & 0\\ 0 & \color{orange}{2} & 0 & 0 & \color{orange}{1} & 0\\ \color{red}{2} & 0 & 0 & 0 & 0 & \color{red}{1}\end{bmatrix} = \det \begin{bmatrix} \color{red}{1} & \color{red}{2} & & & & \\ \color{red}{2} & \color{red}{1} & & & & \\ & & \color{orange}{1} & \color{orange}{2} & & \\ & & \color{orange}{2} & \color{orange}{1} & & \\ & & & & \color{magenta}{1} & \color{magenta}{2} \\ & & & & \color{magenta}{2} & \color{magenta}{1} \end{bmatrix} = \left( \det \begin{bmatrix} 1 & 2\\ 2 & 1\end{bmatrix} \right)^3 = (-3)^3 = \color{blue}{-27}$$

matrices block-matrices permutation-matrices determinant
A: Let ${\rm R}_3$ be the $3 \times 3$ reversal matrix. Hence,
$$\det \left[\begin{array}{ccc|ccc} 1&0&0&0&0&2\\ 0&1&0&0&2&0\\ 0&0&1&2&0&0\\ \hline 0&0&2&1&0&0\\0&2&0&0&1&0\\ 2&0&0&0&0&1\end{array}\right] = \det \begin{bmatrix} {\rm I}_3 & 2{\rm R}_3\\ 2{\rm R}_3 & {\rm I}_3\end{bmatrix} = \det \left( {\rm I}_3 - 4 {\rm R}_3^2 \right) = (-3)^3 = \color{blue}{-27}$$
because ${\rm R}_3^2 = {\rm I}_3$.

matrices block-matrices permutation-matrices determinant
A: As, each row sum is $3$ , so, $3$ is an eigenvalue of the matrix $A$(namely). As , dimension of nullspace of $(A-3I)$ is $3$, Geometric multiplicity of eigenvalue $3$ is $3$.
As, the matrix $A $ is symmetric, so, $A$ is diagonalizable.
Hence, Algebraic multiplicity of eigenvalue of $A$= Geometric multiplicity of eigenvalue of $A$.
So, Algebraic multiplicity of eigenvalue $3$ is $3$.
Now , as, dimension of nullspace of $(A+1I) $ is $3$.
So, $-1$ is another eigenvalue of $A$ with geometric multiplicity $3$.
So, Algebraic multiplicity of eigenvalue $-1$ is $3$.
Now, det($A$)=multiplication of eigenvalues=$(-1)×(-1)×(-1)×3×3×3=-27$
