Set up your matrix from the coefficients of the system.
\(\displaystyle \begin{bmatrix}1&2&5&16\\0&4&4&8\\0&0&1&5\end{bmatrix}\)
The idea is to perform row operations until you get it into the form:
\(\displaystyle \begin{bmatrix}1&0&0&a\\0&1&0&b\\0&0&1&c\end{bmatrix}\)
where a,b,c are your solutions, x=a, y=b, z=c.
On thing nice about this one is that it is already finished to a point.
Multiply the second row by 1/4:
\(\displaystyle \begin{bmatrix}1&2&5&16\\0&1&1&2\\0&0&1&5\end{bmatrix}\)
Multiply the third row by -1 and add to the second row:
\(\displaystyle \begin{bmatrix}1&2&5&16\\0&1&0&-3\\0&0&1&5\end{bmatrix}\)
Multiply the second row by -2 and add to the first row:
\(\displaystyle \begin{bmatrix}1&0&5&22\\0&1&0&-3\\0&0&1&5\end{bmatrix}\)
Multiply the third row by -5 and add to the first row:
\(\displaystyle \begin{bmatrix}1&0&0&-3\\0&1&0&-3\\0&0&1&5\end{bmatrix}\)
There ya' go. x=-3, y=-3, z=5
There was a nice stepped through example. Now can you proceed with others?. These take practice. I know they're laborious and tedious.
A lot of calculators do them. I would use that to check anyway.
