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January 31, 2015

January 31, 2015

Posted by **Becca** on Thursday, March 28, 2013 at 7:42am.

-x-y+z=1

x-y-4z=-7

4x+y+z=6

2.) Perform the indicated row operation, then write the new matrix.

-4 4 | 4

9 -6 | 5 -2R1+ R2 ---> R2

3.) 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.

-x+y+z=0

3x+2y+5z=18

15x+10y+25z=89

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