Question

solve using row reduction:

1. x-3y+z=11
3x+y-2z=-2
x-y-z=-3

2. solve using row reduction

x-2y-3z=-5
2x-y+2z=11
-3x+0y+z=-3

Answer 1

Let's do these one at a time. So let's do #1 first

x-3y+z=11
3x+y-2z=-2
x-y-z=-3



1 -3 1 11
3 1 -2 -2
1 -1 -1 -3



1 -3 1 11
0 10 -5 -35 R2 - 3R1 ---> R2
1 -1 -1 -3



1 -3 1 11
0 10 -5 -35
0 2 -2 -14 R3 - 1*R1 ---> R3



1 -3 1 11
0 1 -5/10 -35/10 (1/10)*R2 ---> R2
0 2 -2 -14



1 -3 1 11
0 1 -1/2 -7/2
0 2 -2 -14



1 -3 1 11
0 1 -1/2 -7/2
0 0 -1 -7 R3 - 2*R2 ---> R3



1 -3 1 11
0 1 -1/2 -7/2
0 0 1 7 -1*R3 ---> R3



1 -3 1 11
0 1 0 0 R2 + (1/2)*R3 ---> R2
0 0 1 7



1 -3 0 4 R1 - 1*R3 ---> R1
0 1 0 0
0 0 1 7



1 0 0 4 R1 + 3*R2 ---> R1
0 1 0 0
0 0 1 7


So the solutions are x = 4, y = 0, z = 7

Answer 2

what did you do for the first step to change the second row?

Answer 3

I subtracted 3 times row 1 from row 2 (and replaced row 2 with this result)

Answer 4

ok i see what youre doing now this can be lengthy

Answer 5

x-2y-3z=-5
2x-y+2z=11
-3x+0y+z=-3



1 -2 -3 -5
2 -1 2 11
-3 0 1 -3



1 -2 -3 -5
0 3 8 21 R2 - 2*R1 ---> R2
-3 0 1 -3



1 -2 -3 -5
0 1 8/3 7 (1/3)*R2 ---> R2
-3 0 1 -3



1 -2 -3 -5
0 1 8/3 7
0 -6 -8 -18 R3 + 3*R1 ---> R3



1 -2 -3 -5
0 1 8/3 7
0 0 8 24 R3 + 6*R2 ---> R3



1 -2 -3 -5
0 1 8/3 7
0 0 1 3 (1/8)*R3 ---> R3



1 -2 -3 -5
0 1 0 -1 R2 - (8/3)*R3 ---> R2
0 0 1 3



1 -2 0 4 R1 +3*R3 ---> R1
0 1 0 -1
0 0 1 3



1 0 0 2 R1 + 2*R2 ---> R1
0 1 0 -1
0 0 1 3



So x = 2, y = -1, and z = 5



Yes they do get quite lengthy, but it's really the best way solve systems of equations.

Answer 6

oops made a typo lol, at the end it should say "So x = 2, y = -1, and z = 3"

Other than that, everything else should be ok.

Answer 7

ok thank-you very much you've been a great help

Answer 8

you're welcome

Answer 9

i wish you could be my math tutor thanks so much now i better understand this

Answer 10

glad I could help you out