# equation

Document Sample

```					Solving Simultaneous Equations Using Row Reduction and a Matrix Approach

Example

Step 1: Identify the system of equations                     x+y+z=6
x–y+z=2
3x + y – z = 2

Step 2: Write in matrix notation                                    x            y             z             b
(Create one column for each variable and one column                1            1             1             6
for the right hand sides—referred to as b.) Don’t worry             1           -1             1             2
about where the box around the 1 comes from at this                 3            1            -1             2
point—that will come in Steps 3 and 4.

Step 3: Select a column to work on (The object is to get     Column 1 (Note: I could choose column 1, 2, or 3)
a unit vector in each column, and to have a one in each
row. Unit vectors will have one one in the column and
everything else in that column will be zero.)

Step 4: Select a row to make into the pivot row.             Row 1 (Note: Although I could select any row, Rows 1
(With steps 3 and 4 you will have identified a specific      or 2 look easiest since the entry in the selected column is
entry—call this a.)                                          already a 1.)

Step 5: To get a one in that position, multiply every        a=1 so 1/1 = 1 and the row remains unchanged.
number in this row by 1/a                                           1            1           1           6
(Call the resulting row the pivot row.)

Step 6: You want to make the other entries in that           Working on the second row:
column zero. To do this, work one row at a time. Select       Multiply the pivot row by –1 gives:
a row. Look at the current entry. Multiply each number             -1            -1          -1             -6
in the pivot row by –1 times this entry and add the result    Adding this to the current row 2 of:
to the corresponding entry in the target row. Repeat this           1            -1           1              2
process until you have finished with the column.              Results in the following new row 2:
0            -2           0             -4

Working on the third row:
Multiply the pivot row by –3 and add:
-3(1)+3      -3(1)+1     -3(1)+(-1)      -3(6)+2
0           -2           -4            -16

At this point we have completed one pivot and we have:
1           1           1             6
0          -2           0            -4
0          -2          -4           -16
Step 7: If you have additional columns to transform into   Working on the second column and selecting the second
unit vectors, return to Step 3.                            row, make this entry a 1 by multiplying the row by –1/2.

0            1             0              2
Call this the pivot row.

Now to make the second entry in the first row a 0,
multiply the pivot row by –1 and add:
-1(0)+1        -1(1)+1      -1(0)+1       -1(2)+6
1            0            1             4

To transform the third row, multiply the pivot row by 2
2(0)+0      2(1)+(-2)      2(0)+(-4)     2(2)+(-16)
0            0           -4           -12

At this point we have completed the second pivot and
we have:
1           0            1            4
0           1            0            2
0           0          -4          -12

For the next pivot, we must get a one in the third row
and third column. We do this by multiplying by –1/4.

0            0             1              3
Call this the pivot row.

The second row already has a zero in the third column,
so the only other work is to get a 0 in the first row third
column. Multiply the pivot row by –1 and add:
-1(0)+1     -1(0)+0         -1(1)+1       -1(3)+4
1           0              0               1

After three pivots we have:
x           y              z
1           0              0              1
0           1              0              2
0           0              1              3

1            0             0              1
0            1             0              2
0            0             1              3