Posted by **Kris** on Thursday, March 28, 2013 at 11:29am.

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.) (1,1,1)

C.) Coincident dependence; (x,y,z)|3x+yz=4}

D.) No solution, inconsistent

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

- Algebra Multiple choice revised!! HELP -
**Steve**, Thursday, March 28, 2013 at 1:43pm
#1 - look at the equations. They are all the same. That is, they are multiples of each other. So, (C)

#2. (D) Did you check your answer to see whether it actually fits the equations?

#3. (A) and (D) are the same, so I suspect a typo. Anyway, you want to get rid of the 8, so you know you have to multiply R1 by 8 and subtract R2.

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