Question

I have a few questions I need help with! Please explain if my answer was not right how you got to it. I got 1.) B 2.) A 3.) C

1. Solve the system by triangularizing the augmented matrix and using back substitution. If the system is linearly dependent, give the solution in terms of a parameter.
-3x+y-z=-4
6x-2y+2z=8
-12x+4y-4z=-16
A.)(1,1,2)
B.)(0, -2)
C.) (0,-3)
D.) (-1,-2)

2. Solve the system by triangularizing the augmented matrix and using back substitution.
6x=y-4
3y=-13-7x
a.) (-1,-3)
b.) (0,-2)
C.) (0,-3)
D.) (-1,-2)

3. What row operation would produce zero beneath the first entry in the diagonal?
1 2 | -1
8 -2 | -2
A.) 8R1-R2--> R2
B.) -8R1-R2 -->R2
C.) -R1+R2 --> R2
D.) 8R1-R2 --> R2

Answer 1

for 2 answer is <d>
1 is ok