Matrices and Systems of Equations

A matrix derived from a system of linear equations is the augmented matrix of the system.

Example:

2x + y - 3z = 4
x + 2y - z = 10
3x - 2y + 2z = 5

| 2   1  -3    4|
| 1   2  -1  10|
| 3  -2   2    5|

The first three columns is the matrix of the coefficients of x, y, and z.  The last column is the right side of the system of equations.

Elementary Row Operations:
1.  Interchange two rows.
2.  Multiply a row by a nonzero constant.
3.  Add a multiple of a row to another row.

| 2   1  -3    4|
| 1   2  -1  10|
| 3  -2   2    5|

Work column by column.
 

Column 1:

Multiple Row 1 by 1/2 to get a 1 in the first position.

| 1   1/2  -3/2    2|
| 1      2     -1  10|
| 3     -2      2    5|

Now, put zeros in the two positions below the 1 by multiplying Row 1 by something and adding it to Row 2 or Row 3.

    Multiply Row 1 by -1 and add it to Row 2 and put the answer in Row 2.

    | 1   1/2  -3/2    2|
    | 0    3/2  1/2    0|
    | 3     -2      2    5|

    Multiply Row 1 by -3 and add it to Row 3 and put the answer in Row 3.

     | 1   1/2  -3/2     2|
     | 0    3/2  1/2     0|
     | 0  -7/2  13/2  -1|
 

Column 2:

Now, work on column 2 and put a 1 in the second position in the second column by multiplying Row 2 by 2/3.

 | 1     1/2    -3/2          2|
 | 0         1    1/3     16/3|
 | 0    -7/2    13/2       -1|

And, now, put 0's in the first and third rows above and below that 1 by taking appropriate actions.

Multiply Row 2 by -1/2 , add it to Row 1, and put the answer in Row 1.

| 1          0   -5/3     -2/3|
| 0         1    1/3     16/3|
| 0    -7/2    13/2       -1|

Multiply Row 2 by 7/2, add it to Row 3, and put the answer in Row 3.

| 1          0   -5/3     -2/3|
| 0         1    1/3     16/3|
| 0         0    23/3   53/3|
 

Column 3:

Now, work on the third column, making a 1 in the third row, third position.

Multiply Row 3 by 3/23.

| 1          0   -5/3      -2/3|
| 0         1    1/3      16/3|
| 0         0      1     53/23|

The last thing to do is to make 0's appear in the first and second rows above the 1 in the third row.

Multiply Row 3 by 5/3 and add it to Row 1, putting the answer in Row 1.

| 1          0      0    73/23|
| 0         1    1/3      16/3|
| 0         0      1     53/23|

Multiply Row 3 by -1/3, add it to Row 2, and put the answer in Row 2.

| 1          0      0    73/23|
| 0         1       0  105/23|
| 0         0      1     53/23|

This is in row-reduced form and essentially says  x = 73/23, y = 105/23, and z = 53/23.
 
 


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