## anonymous one year ago Find all solutions of the congruences $x+y+z\equiv 1 (mod 7)$ $x+y+w\equiv 1 (mod 7)$ $x+z+w\equiv 1 (mod 7)$ $y+z+w\equiv 1 (mod 7)$

1. anonymous

\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*}

2. anonymous

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

3. anonymous

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.