Finding determinant of following matrix

Question

I need to find determinant of following matrix.

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

Original Image

I did it by simply doing $R_5$ - $R_1$. and then evaluating the determinant. But its a lengthy process but answer came out..

But another thing i have noticed afterwards is that it is symmetric matrix. So finding determinant my be easy, but i don't know how it helps which is my main concern. Thanks

Answer 1

Might be easiest to row-reduce just enough so that the matrix becomes a triangular matrix. Then the determinant is the product of the diagonal entries. In this case, we can row reduce to get $$\begin{bmatrix} 1 & 0 & 0 & 0 &0 & 2\\0& 1&0& 2 & 2 &0\\0 & 0 & 1&2&0&0\\0&0&0&-3&0&0\\0&0&0&0&-3&0\\0&0&0&0&0&-3\end{bmatrix},$$ and so the determinant is $-27$.

Answer 2

Why not $\;R_5-2R_1\;$ ? Adding scalar multiples of one row to other one doesn't change the determinant. Then you can develop wrt the first column:

$$\begin{vmatrix} 1&0&0&0&0&2\\ 0&1&0&2&2&0\\ 0&0&1&2&0&0\\ 0&0&2&1&0&0\\ 0&2&0&0&1&0\\ 2&0&0&0&0&1\end{vmatrix}\stackrel{R_5-2R_1}\longrightarrow\begin{vmatrix} 1&0&0&0&0&2\\ 0&1&0&2&2&0\\ 0&0&1&2&0&0\\ 0&0&2&1&0&0\\ 0&2&0&0&1&0\\ 0&0&0&0&0&\!\!-3\end{vmatrix}=1\begin{vmatrix} 1&0&2&2&0\\ 0&1&2&0&0\\ 0&2&1&0&0\\ 2&0&0&1&0\\ 0&0&0&0&\!\!-3\end{vmatrix}=$$

$$=-3\begin{vmatrix} 1&0&2&2\\ 0&1&2&0\\ 0&2&1&0\\ 2&0&0&1\\\end{vmatrix}=\ldots etc.$$