finding a smart way to calculate a determinant

Question

I was trying to solve the following determinant: $$det(M)=\begin{vmatrix}2 & -1 & -1 & 0 & 0 & 0 & 0\\ 0 & 2 & 0 & -1 & -1 & 0 & 0\\ 0 & 0 & 2 & 0 & 0 & -1 & -1\\ -1 & 0 & 0 & 2 & 0 & 0 & 0\\ 0 & -1 & -1 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & -1 & -1 & 2 & 0\\ 0 & 0 & 0 & 0 & 0 & -1 & 1 \end{vmatrix}$$

I have succeeded in solving it and proving that $det(M) = 16$, but It took me a while and I just moved from one minor into two and so on. I was wondering if there is a "smart" way to solve this determinant. For example, so how to get in into a triangular matrix and then just to multiply the diagonal.

Answer 1

A smart way would be using the Elementary Row or Column Operations to generate more zero's.

For example if you perform $$ C_6+C_7\to C_6 $$ you will generate a column with one more zero.

Answer 2

You can do multiple things that will make the determinant unchanged yet the matrix will be much more simple. Amongst those that leaves the determinant invariant are the following.

1. Adding arbitrary multiple of one row to another row.
2. Adding arbitrary multiple of one column to another column.
3. Transposing the matrix.

There are other rules you can use that simplifies the matrix but changes the determinant so you have to be careful to count in the changes, e.g.

1. Multiplying a row/column with a number yields a matrix that has the determinant of the original matrix times the number you multiplied the row/column.
2. Switching rows/columns changes the sign of the determinant.

There are also other rules you can apply, more information you can find, for example in this wiki article about determinant. Note that you can combine all the above steps so you get quite a toolkit of possible steps to simplify your matrix before actually computing the determinant.

Answer 3

It can be obtained in $8$ strokes (just like The Hunting of the Snark!):

Observe first that if you add all columns, all rows will be $0$ but the fourth. So we'll begin with adding columns $2$ to $7$ to the first, and expand along the first column:: \begin{align} &\begin{vmatrix} 0&-1&-1&0&0&0&0 \\ 0&2&0&-1&-1&0&0 \\ 0&0&2&0&0&-1&-1 \\ 1&0&0&2&0&0&0 \\ 0&-1&-1&0&2&0&0 \\ 0&0&0&-1&-1&2&0 \\ 0&0&0&0&0&-1&1 \end{vmatrix} =-\begin{vmatrix} -1&-1&0&0&0&0 \\ 2&0&-1&-1&0&0 \\ 0&2&0&0&-1&-1 \\ -1&-1&0&2&0&0 \\ 0&0&-1&-1&2&0 \\ 0&0&0&0&-1&1 \end{vmatrix} \\[2ex] =-&\begin{vmatrix} -2&-1&0&0&0&0 \\ 0&0&-1&-1&0&0 \\ 0&2&0&0&-1&-1 \\ 0&-1&0&2&0&0 \\ 0&0&-1&-1&2&0 \\ 0&0&0&0&-1&1 \end{vmatrix} =2\begin{vmatrix} 0&-1&-1&0&0 \\ 2&0&0&-1&-1 \\ -1&0&2&0&0 \\ 0&-1&-1&2&0 \\ 0&0&0&-1&1 \end{vmatrix} =2\begin{vmatrix} 0&-1&-1&0&0 \\ 2&0&0&-2&0 \\ -1&0&2&0&0 \\ 0&-1&-1&2&0 \\ 0&0&0&-1&1 \end{vmatrix}\\[2ex] \overset{\begin{array}{l}\text{(expanding along}\\ \text{the last column)}\end{array}}{=} 2\,&\begin{vmatrix} 0&-1&-1&0 \\ 2&0&0&-2 \\ -1&0&2&0 \\ 0&-1&-1&2 \end{vmatrix}=2 \begin{vmatrix} 0&-1&-1&0 \\ 2&-1&-1&0 \\ -1&0&2&0 \\ 0&-1&-1&2 \end{vmatrix} \overset{\begin{array}{l}\text{(expanding along}\\ \text{the last column)}\end{array}}{=} 4\begin{vmatrix} 0&-1&-1 \\ 2&-1&-1 \\ -1&0&2 \end{vmatrix}\\[2ex] =4\,&\begin{vmatrix} 0&-1&0 \\ 2&-1&0 \\ -1&0&2 \end{vmatrix}=8\begin{vmatrix} 0&-1 \\ 2&-1 \end{vmatrix}=16. \end{align}