# Algebra Multiple choice revised!! HELP

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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 -

#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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