System of linear equations over field

Question

Solve the system of linear equations $$ \begin{split} 2&x\ &+\ 17&y\ &= 1\\ 7&x\ &+\ &y\ &= 2 \end{split} $$ over $\mathbb{Z}_{23}$.

Can someone explain to me how to do this?

Answer 1

Use row reduction with the augmented matrix:

\begin{align}% &\begin{bmatrix}\mkern-8mu\begin{array}{[rr|r} 2&17&1\cr 7&1&2 \end{array}\mkern-8mu\end{bmatrix}\xrightarrow{R_1\leftarrow12R_1} % \begin{bmatrix}\mkern-8mu\begin{array}{[rr|r} 1&-3&12\cr 7&1&2 \end{array}\mkern-8mu\end{bmatrix}\xrightarrow{R_2\leftarrow 7R_1-R_2} % \begin{bmatrix}\mkern-8mu\begin{array}{[rr|r} 1&-3&12\cr 0&1&13 \end{array}\mkern-8mu\end{bmatrix} \\[1ex]&\xrightarrow{R_1\leftarrow R_1+3R_2} % \begin{bmatrix}\mkern-8mu\begin{array}{[rr|r} 1&0&\color{red}5\cr 0&1&\color{red}{13} \end{array}\mkern-8mu\end{bmatrix} \end{align}

Answer 2

You do it exactly the same like in the Rational numbers, just obey the arithmetic in $\Bbb Z/23\Bbb Z$.

For example, $\frac12\equiv12 \pmod{23}$ because $2\cdot 12=24\equiv1\pmod{23}.$

Answer 3

From the second equation we obtain: $$6(7x+y)=12$$ or $$42x+6y=12,$$ which gives $$44x=13$$ or $$-2x\equiv 13$$ or $$x\equiv11\cdot13$$ or $$x\equiv5.$$ Can you end it now?