Question

For the linear system with augmented matrix shown below, find the REF and then solve the system by back substitution. (Do not do any row exchanges. And do not use the ref command on your calculator because the calculator uses row exchanges and will?

For the linear system with augmented matrix shown below, find the REF and then solve the system by back substitution. (Do not do any row exchanges. And do not use the ref command on your calculator because the calculator uses row exchanges and will generally give a different REF than is expected.)

4 −8 8 −16
8 −13 19 −23
12 −36 9 −75

Answer 1

See below for step by step breakdown.

#[(1, -2, 2 | -4), (0, 1, 1 | 3), (0, 0, 1 | -3)]#

Explanation:

#[(4, -8, 8 | -16), (8, -13, 19 | -23), (12, -36, 9 | -75)]#

To reduce this matrix, we will first perform the row operations `R_2 - 2R_1` and `R_3-3R_1`, followed by `1/4 R_1`

#[(4, -8, 8 | -16), (8, -13, 19 | -23), (12, -36, 9 | -75)]#

`R_2-2R_1 & R_3-3R_1 ->`

#[(4, -8, 8 | -16), (0, 3, 3 | 9), (0, -12, -15 | -27)]#

#1/4 R1 -> [(1, -2, 2 | -4), (0, 3, 3 | 9), (0, -12, -15 | -27)]#

It then behooves us to perform the operation `R_3 + 4R_2` before performing `1/3 R_2`...

#R_3+4R_2 -> [(1, -2, 2 | -4), (0, 3, 3 | 9), (0, 0, -3 | 9)]#

And now we will perform `1/3 R_2 & -1/3 R_3`...
#1/3 R_2 & -1/3 R_3 -> [(1, -2, 2 | -4), (0, 1, 1 | 3), (0, 0, 1 | -3)]#