Linear combination aka elimination method
\begin{align} 2x &+ 3y &= 5 \\ x &+ y &= 1 \end{align}is equivalent to
\[ \left[\begin{matrix} 2 &3 \\ 1 &1 \end{matrix}\right] \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}5\\1\end{matrix}\right] \].
We want to multiply the bottom equation by \(-2\) when solving with the elimination method normally, so we might expect to multiply by the identity matrix but with the "bottom row selector" modified:
\(\left[\begin{matrix}1&0\\0&-2\end{matrix}\right]\).
\[ \left[\begin{matrix} 1 &0 \\ 0 &-2\end{matrix}\right] \left[\begin{matrix} 2 &3 \\ 1 &1 \end{matrix}\right] \left[\begin{matrix} x \\ y \end{matrix}\right] \]
=
\[ \left[\begin{matrix} 1 &0 \\ 0 &-2\end{matrix}\right] \left[\begin{matrix} 5 \\ 1\end{matrix}\right] \]
And then, to add the bottom to the top we can use \(\left[\begin{matrix}1&1\\0&1\end{matrix}\right]\).
KBe2020math530srcMatriciesAsEquationsIntro.png KBe2020math530floMatrixMultiplyToSolve.png
