How do you solve #3x + 2y + 2z =3#, #x + 2y - z =5# and #2x - 4y + z=0# using matrices?
1 Answer
The answer is
Explanation:
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|