# 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:

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

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

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

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

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

Divide all entries in 2nd row by -15.

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

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

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

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

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

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

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

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

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

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:

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

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