Question

Matrix help?

x+y-9z=9
9x-y-18z=-2

that is the system of equations, and I got {(7+27z)/10, (83-63z)/10, z}, but it's telling me that the "y" is wrong. . . what did I do wrong?

Answer 1

Eliminating \(x\) in the second equation (by taking \(9\text{ row}_1-\text{row}_2\) to replace \(\text{row}_2\)) gives
\[\left[\begin{array}{ccc|c}
1&1&-9&9\\[1ex]
9&-1&-18&-2
\end{array}\right]

\implies

\left[\begin{array}{ccc|c}
1&1&-9&9\\[1ex]
0&-10&63&-83
\end{array}\right]\]Next, eliminating \(y\) in the first row (by taking \(10\text{ row}_1+\text{row}_2\) to replace \(\text{row}_1\)) gives
\[\left[\begin{array}{ccc|c}
1&1&-9&9\\[1ex]
0&-10&63&-83
\end{array}\right]

\implies

\left[\begin{array}{ccc|c}
10&0&-27&7\\[1ex]
0&-10&63&-83
\end{array}\right]\]Divide through by \(10\) in the first row and \(-10\) in the second to get the fully row-reduced form
\[\left[\begin{array}{ccc|c}
1&0&-\dfrac{27}{10}&\dfrac{7}{10}\\[1ex]
0&1&-\dfrac{63}{10}&\dfrac{83}{10}
\end{array}\right]\]
Fixing \(z\), the second equation tells you that
\[y-\frac{63}{10}z=\frac{83}{10}\implies y=\frac{63z+83}{10}\]