Thread:

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:

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

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

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

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

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

Divide all entries in 2nd row by -15.

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

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

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

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

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

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

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

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

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

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:

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

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