Algebra Multiple choice revised!! HELP
posted by Kris .
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+yz=4
6x2y+2z=8
12x+4y4z=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=y4
3y=137x
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.) 8R1R2> R2
B.) 8R1R2 >R2
C.) R1+R2 > R2
D.) 8R1R2 > R2

#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.