Linear system $AX=0$ has a nonzero solution

Question

Which parameters $a,b,c,d$ satisfy the matrix $$A=\begin{bmatrix}-1 & 1&1&1 \\1 & -1&1&1\\1 &1 &-1 &1\\a&b&c&d \end{bmatrix}$$ sucht that linear system $AX=0$ has a nonzero solution?

Again by row reduing the matrix to echelon form isn't it?

Answer 1

$$A=\begin{pmatrix}-1 & 1&1&1 \\1 & -1&1&1\\1 &1 &-1 &1\\a&b&c&d \end{pmatrix}\longrightarrow\begin{pmatrix}-1 & 1&1&1 \\0 & 0&2&2\\0 &2 &0 &2\\0&b+a&c+a&d+a \end{pmatrix}\longrightarrow$$

$${}$$

$$\begin{pmatrix}-1 & 1&1&1 \\0 & 1&0&1\\0 &0 &2 &2\\0&b+a&c+a&d+a \end{pmatrix}\longrightarrow\begin{pmatrix}-1 & 1&1&1 \\0 & 1&0&1\\0 &0 &2 &2\\0&0&c+a&d-b \end{pmatrix}$$

End the argument.

Answer 2

According to entries at the first $3$ rows, if we sum up the first three columns we will get the entry in the fourth column, since $$-1+1+1=1$$ Thus any solution that satisfies $$a+b+c=d$$ yields a linearly dependent matrix (where the last column can be written as the sum of the first three columns).

Answer 3

Similar to Stefanos's answer

The matrix has determinant $$ \det(A) = \det \begin{pmatrix}-1 & 1&1&1 \\1 & -1&1&1\\1 &1 &-1 &1\\a&b&c&d \end{pmatrix} = 4(d-c-b-a) $$

for calculating the determinant see, http://en.wikipedia.org/wiki/Laplace_expansion.

The linear system has only the zero solution if $(d-c-b-a)\ne 0$.

Thus for a nonzero solution one needs $a+b+c=d$.