<s>[INST] 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.
 [/INST] 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 $\det(A_n)=(-3)^n$.
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