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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} \begin{pmatrix}1&1&1&0\\ 1&1&0&1\\ 1&0&1&1\\ 0&1&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\\w\end{pmatrix}&\equiv\begin{pmatrix}1\mod7\\1\mod7\\1\mod7\\1\mod7\end{pmatrix}\\[2ex] \begin{pmatrix}x\\y\\z\\w\end{pmatrix}&\equiv \frac{1}{3}\begin{pmatrix}1&1&1&2\\1&1&2&1\\1&2&1&1\\2&1&1&1\end{pmatrix} \begin{pmatrix}1\mod7\\1\mod7\\1\mod7\\1\mod7\end{pmatrix} \end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles Would you please explain to me how you got the second line. I am so confused. Thanks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Assuming 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.
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