How do you solve #3x + 2y + 2z =3#, #x + 2y - z =5# and #2x - 4y + z=0# using matrices?

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Answer 1 · 2016-09-29T23:43:51

The answer is #x = 2, y = 1/2, z = -2#. Please see the explanation for the row operations.

Put the row for x + 2y - z = 5 at the top, because the coefficient of x is 1:

|1 2 -1 |5|

Put the row for 3x + 2y + 2z = 3 next:

|1 2 -1 |5|
|3 2 2|3|

Put the row for 2x -4y + z = last:

|1 2 -1 |5|
|3 2 2|3|
|2 -4 1|0|

Multiply row 1 by -3 and add to row 2:

|1 2 -1 |5|
|0 -4 5|-12|
|2 -4 1|0|

Multiply row 1 by -2 and add to row 3:

|1 2 -1 |5|
|0 -4 5|-12|
|0 -8 3|-10|

Multiply row 2 by -2 and add to row 3:

|1 2 -1 |5|
|0 -4 5|-12|
|0 0 -7|14|

Divide row 3 by -7

|1 2 -1 |5|
|0 -4 5|-12|
|0 0 1|-2|

Multiply row 3 by -5 and add to row 2:

|1 2 -1 |5|
|0 -4 0|-2|
|0 0 1|-2|

Divide row 2 by -4:

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

Add row 3 to row 1:

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

Multiply row 2 by -2 and add to row 1:

|1 0 0|2|
|0 1 0|1/2|
|0 0 1|-2|