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I just want to check that I did a certain problem correctly. This is it: $$a+b=3 \pmod{26}\\2a+b=7 \pmod{26}$$ Solve for $a$ and $b$

Now I setup the augmented matrix: $$\left[ \begin{array}{ccc} 1 & 1 & 3 \\ 2 & 1 & 7 \end{array} \right]$$

After doing elementary row operations I get: $$\left[ \begin{array}{ccc} 1 & 0 & 4 \\ 0 & 1 & -1 \end{array} \right]$$

That yields: $$4-1=3\pmod{26}\\ 8-1=7\pmod{26}$$

Did I do that correctly?

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Yes you did ${}{}{}{}$ – Amr Jan 21 '13 at 19:42
@CharlieYabben : If you look at my edits, you will see how \pmod is used and how "displaye" TeX is done. – Michael Hardy Jan 21 '13 at 22:19
up vote 2 down vote accepted

The matrix manipulations are correct, but the interpretation of the answer isn't. After doing row operations, your matrix is the augmented matrix of the system \begin{align*} a \qquad & \equiv \phantom{-}4 \pmod{26} \\ \qquad b & \equiv -1 \pmod{26}. \end{align*} So that's your solution: $a\equiv4\pmod{26}$, $b\equiv-1\equiv25\pmod{26}$. You can check that this solution satisfies both original congruences (and also check that the solution you originally obtained does not).

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