# Thread: Matrix- using row operations

1. ## Matrix- using row operations

1.) 7x + 5y - z = 94
x + 5y + 2z = 52
2x + y + z = 29

2. Originally Posted by kelsikels
1.) 7x + 5y - z = 94
x + 5y + 2z = 52
2x + y + z = 29

Set up:

$
\left(\begin{array}{ccc|c}
7 & 5 & -1 & 94 \\
1 & 5 & 2 & 52 \\
2 & 1 & 1 & 29
\end{array}\right)
$

Move the 2nd row to the top row. You'll see why.

$
\left(\begin{array}{ccc|c}
1 & 5 & 2 & 52 \\
7 & 5 & -1 & 94 \\
2 & 1 & 1 & 29
\end{array}\right)
$

Take the 2nd row: subtract 7x the top row from each entry.

$
\left(\begin{array}{ccc|c}
1 & 5 & 2 & 52 \\
0 & -30 & -15 & -270 \\
2 & 1 & 1 & 29
\end{array}\right)
$

Divide all entries in 2nd row by -15.

$
\left(\begin{array}{ccc|c}
1 & 5 & 2 & 52 \\
0 & 2 & 1 & 18 \\
2 & 1 & 1 & 29
\end{array}\right)
$

Third row: subtract 2x the top row from each entry.

$
\left(\begin{array}{ccc|c}
1 & 5 & 2 & 52 \\
0 & 2 & 1 & 18 \\
0 & -9 & -3 & -75
\end{array}\right)
$

Divide each entry in the 3rd row by -3.

$
\left(\begin{array}{ccc|c}
1 & 5 & 2 & 52 \\
0 & 2 & 1 & 18 \\
0 & 3 & 1 & 25
\end{array}\right)
$

Third row: subtract 3/2x of the 2nd row.

$
\left(\begin{array}{ccc|c}
1 & 5 & 2 & 52 \\
0 & 2 & 1 & 18 \\
0 & 0 & -\frac{1}{2} & -2
\end{array}\right)
$

Multiply the entries in the 3rd row by -2.

$
\left(\begin{array}{ccc|c}
1 & 5 & 2 & 52 \\
0 & 2 & 1 & 18 \\
0 & 0 & 1 & 4
\end{array}\right)
$

This can be solved by back substitution, now. If you want to practice your row skills, find the answer by back substitution, then check yourself: find the answer directly by reducing it to this form:

$
\left(\begin{array}{ccc|c}
1 & 0& 0 & x \\
0 & 1 & 0 & y \\
0 & 0 & 1 & z
\end{array}\right)
$

You might also do the steps above yourself and check my work. Maybe I tossed in a monkey wrench somewhere along the way ...?