Consider the matrix $M=$
\begin{bmatrix}
1&1&1&1&1&1&1\\
1&-1&1&1&1&0&0\\
1&1&-1&1&1&0&0\\
1&1&1&-1&1&0&0\\
1&1&1&1&-1&0&0\\
1&0&0&0&0&-3&1\\
1&0&0&0&0&1&-3\\
\end{bmatrix}
Show that $M$ has determinant $0$ and its rank is $6$.
In order to show that determinant of the matrix is $0$ we need to expand $M$ along the last row by Laplace Expansion which is cumbersome.
Are there any other methods to show that $M$ has determinant zero.
I have shown the second part that rank(M)=6 because we have the sub-matrix
\begin{bmatrix}
-1&1&1&1&0&0\\
1&-1&1&1&0&0\\
1&1&-1&1&0&0\\
1&1&1&-1&0&0\\
0&0&0&0&-3&1\\
0&0&0&0&1&-3\\
\end{bmatrix}
which is block diagonal matrix and each of its block has non-zero determinant.
However how to show the first part? Is there any elegant way to show this as the matrix has a definite pattern?
Please help.
Best Answer
UPDATE:
After perforoming a row operation, it is easy for me to spot that $$ (2, -1,-1,-1,-1,1,1) $$ is a nontirivial solution as @user1551 commented. Thanks for that hint.
Disclaimer: this method might not be elegant enough as you wanted.
Solution. $\blacktriangleleft$ Add the $(-1)$ $\times$ 1st row to rows 2 thru 5, we have $$ \det (\boldsymbol M) =: A = \begin{vmatrix} 1&1&1&1&1&1&1\\ 0&-2 &0 & 0 & 0 & -1 &-1\\ 0& 0&-2 & 0 & 0 & -1 &-1\\ 0& 0 &0 & -2 & 0 & -1 &-1\\ 0& 0 &0 & 0 & -2 & -1 &-1\\ 1& 0 &0 & 0 & 0 & -3 & 1\\ 1& 0 &0 & 0 & 0 & 1 & -3 \end{vmatrix}, $$ Now add the $(-1)$ $\times$ 1st column to the 6th and 7th column: $$ A = \begin{vmatrix} 1&1&1&1&1&0&0\\ 0&-2 &0 & 0 & 0 & -1 &-1\\ 0& 0&-2 & 0 & 0 & -1 &-1\\ 0& 0 &0 & -2 & 0 & -1 &-1\\ 0& 0 &0 & 0 & -2 & -1 &-1\\ 1& 0 &0 & 0 & 0 & -4 & 0\\ 1& 0 &0 & 0 & 0 & 0 & -4 \end{vmatrix}. $$ Note that the right lower corner seems triangular, so we expand the determinant along the 1st column, then $A = A_1 - A_6 + A_7$ where the indices indicate the row.
$A_1$ is easy to calculate, since it is triangular, and we have $A_1 = 4^4 = 256$. For $A_6$: $$ A_6 = \begin{vmatrix} 1 & 1 & 1 & 1 & 0 &0\\ -2 &0 & 0 & 0 & -1 &-1\\ 0&-2 & 0 & 0 & -1 &-1\\ 0 &0 & -2 & 0 & -1 &-1\\ 0 &0 & 0 & -2 & -1 &-1\\ 0 &0 & 0 & 0 & 0 & -4 \end{vmatrix} = -4\begin{vmatrix} 1&1&1&1&0\\ -2 &0 & 0 & 0 & -1 \\ 0&-2 & 0 & 0 & -1 \\ 0 &0 & -2 & 0 & -1 \\ 0 &0 & 0 & -2 & -1 \end{vmatrix} $$
Add $1/2$ times rows of no. 2,3,4,5 to the 1st row: $$ A_6 = -4\begin{vmatrix} 0&0&0&0&-2\\ -2 &0 & 0 & 0 & -1 \\ 0&-2 & 0 & 0 & -1 \\ 0 &0 & -2 & 0 & -1 \\ 0 &0 & 0 & -2 & -1 \end{vmatrix} = 8\begin{vmatrix} -2 &0 & 0 & 0\\ 0&-2 & 0 & 0 \\ 0 &0 & -2 & 0\\ 0 &0 & 0 & -2 \end{vmatrix} = 128. $$ For $A_7$, $$ A_7 = \begin{vmatrix} 1 & 1 & 1 & 1 & 0 &0\\ -2 &0 & 0 & 0 & -1 &-1\\ 0&-2 & 0 & 0 & -1 &-1\\ 0 &0 & -2 & 0 & -1 &-1\\ 0 &0 & 0 & -2 & -1 &-1\\ 0 &0 & 0 & 0 & -4 & 0 \end{vmatrix} = 4 \begin{vmatrix} 1 & 1 & 1 & 1 & 0\\ -2 &0 & 0 & 0 & -1\\ 0&-2 & 0 & 0 & -1\\ 0 &0 & -2 & 0 &-1\\ 0 &0 & 0 & -2 &-1 \end{vmatrix} = 4 \times (-32) = -128 $$ as calculated above. Hence $A = 256 - 128 + (-128) = 0$. $\blacktriangleright$