Question

Find all solutions of the congruences
\[x+y+z\equiv 1 (mod 7)\]
\[x+y+w\equiv 1 (mod 7)\]
\[x+z+w\equiv 1 (mod 7)\]
\[y+z+w\equiv 1 (mod 7)\]

Answer 1

\[\begin{align*}
\begin{pmatrix}1&1&1&0\\
1&1&0&1\\
1&0&1&1\\
0&1&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\\w\end{pmatrix}&\equiv\begin{pmatrix}1\mod7\\1\mod7\\1\mod7\\1\mod7\end{pmatrix}\\[2ex]
\begin{pmatrix}x\\y\\z\\w\end{pmatrix}&\equiv
\frac{1}{3}\begin{pmatrix}1&1&1&-2\\1&1&-2&1\\1&-2&1&1\\-2&1&1&1\end{pmatrix}
\begin{pmatrix}1\mod7\\1\mod7\\1\mod7\\1\mod7\end{pmatrix}

\end{align*}\]

Answer 2

@SithsAndGiggles Would you please explain to me how you got the second line. I am so confused. Thanks

Answer 3

Assuming the operations I carried out were valid (I'm not as comfortable with modular arithmetic as I should be), all I did was find the inverse of the coefficient matrix on the LHS, then multiplied both sides by this inverse.
\[\begin{pmatrix}1&1&1&0\\
1&1&0&1\\
1&0&1&1\\
0&1&1&1\end{pmatrix}^{-1}=\frac{1}{3}\begin{pmatrix}1&1&1&-2\\1&1&-2&1\\1&-2&1&1\\-2&1&1&1\end{pmatrix}\]
and for any invertible matrix \(A\) with inverse \(A^{-1}\), we have \(AA^{-1}=A^{-1}A=I\), where \(I\) is the identity matrix.

The actual computation was done by hand and a bit tedious, but you can use a calculator.