Show that the matrix has determinant $=0$

Question

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.

Answer 1

There is no need to do any tedious row/column operations. All you need is some simple inspection.

As $M$ is a $7\times7$ matrix of rank $\ge6$, if $\det M=0$, the null space of $M$ must be one-dimensional.

As you have shown, the last six columns of $M$ are linearly independent. Therefore, if $x=(a,b,c,d,e,f,g)^T\ne0$ lies in the null space of $M$, then $a$ must be nonzero. Also, by symmetry of coefficients in the last two rows/columns of $M$, if $M(a,b,c,d,e,f,g)^T=0$, then $M(a,b,c,d,e,g,f)^T$ must be zero too. It follows that if the null space of $M$ is indeed one-dimensional, we must have $f=g$.

So, by looking at the last two rows of the equation $M(a,b,c,d,e,f,f)^T=0$, we may assume that $f=1$ and $a=2$. The equation $M(2,b,c,d,e,1,1)^T=0$ then reduces to $$ \begin{bmatrix} 1&1&1&1\\ -1&1&1&1\\ 1&-1&1&1\\ 1&1&-1&1\\ 1&1&1&-1\\ \end{bmatrix} \begin{bmatrix}b\\ c\\ d\\ e\end{bmatrix} =\begin{bmatrix}-4\\ -2\\ -2\\ -2\end{bmatrix}, $$ which is clearly solvable by $b=c=d=e=-1$. Hence $M(2,-1,-1,-1,-1,1,1)^T=0$ and in turn $\det M=0$.

Answer 2

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 $A = 256 - 128 + (-128) = 0$. $\blacktriangleright$

Answer 3

Row reduction: $$\begin{vmatrix} 1&1&1&1&1&1&1\\ 1&-1&1&1&1&0&0\\ 1&1&-1&1&1&0&0\\ 1&1&1&-1&1&0&0\\ 1&1&1&1&-1&0&0\\ 1&0&0&0&0&-3&1\\ 1&0&0&0&0&1&-3\\ \end{vmatrix}\stackrel{-C_1+C_6\to C6\\ \ 3C_1+C_7\to C_7}{=} \begin{vmatrix} 1&1&1&1&1&0&4\\ 1&-1&1&1&1&-1&3\\ 1&1&-1&1&1&-1&3\\ 1&1&1&-1&1&-1&3\\ 1&1&1&1&-1&-1&3\\ 1&0&0&0&0&-4&4\\ 1&0&0&0&0&0&0\\ \end{vmatrix}=\\ \begin{vmatrix} 1&1&1&1&0&4\\ -1&1&1&1&-1&3\\ 1&-1&1&1&-1&3\\ 1&1&-1&1&-1&3\\ 1&1&1&-1&-1&3\\ 0&0&0&0&-4&4\\ \end{vmatrix}\stackrel{C_6+C_5\to C_5}{=} \begin{vmatrix} 1&1&1&1&4&4\\ -1&1&1&1&2&3\\ 1&-1&1&1&2&3\\ 1&1&-1&1&2&3\\ 1&1&1&-1&2&3\\ 0&0&0&0&0&4\\ \end{vmatrix}=\\ 4\begin{vmatrix} 1&1&1&1&4\\ -1&1&1&1&2\\ 1&-1&1&1&2\\ 1&1&-1&1&2\\ 1&1&1&-1&2\\ \end{vmatrix}\stackrel{C_1+C_2+C_3+C_4\to C_4}{=} 4\begin{vmatrix} 1&1&1&4&4\\ -1&1&1&2&2\\ 1&-1&1&2&2\\ 1&1&-1&2&2\\ 1&1&1&2&2\\ \end{vmatrix}=0$$