Solve the system of equations using matrices. Use Gaussian elimination with back-substitution. x + y + z = -5 x - y + 3z = -1 4x + y + z = -2

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Solve the system of equations using matrices. Use Gaussian elimination with back-substitution. x + y + z = -5 x - y + 3z = -1 4x + y + z = -2

Mathematics
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[ 1 1 1 ] [ x ] [-5] [ 1 -1 3 ] [ y ] = [-1] [ 4 1 1 ] [ z ] [-2]
yesss ok so what do i do here

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Other answers:

[ 1 1 1 | -5] [ 1 -1 3 |-1] [ 4 1 1 | -2]
we want make this [ .........|..] [ x...... |..] [ ........ |..] element into a zero
take away the first row from the second row
how
[ 1 1 1 | -5] (R1) [ 1 -1 3 | -1] (R2) [ 4 1 1 | -2] (R3) (R2)-(R1) = [(1-1) (-1-1) (3-1) | (-1-5)]
ohhh
whats next or is that it
we get [ 1 1 1 | -5] (R1) [ 0 -2 2 | -6] (R2) [ 4 1 1 | -2] (R3)
now we want make this [ 1.......|..] [ 0...... |..] [ x........ |..] element zero
so take away line row 1 from row 3, four times
soo 111 i think
[ 4 1 1 | -2] - 4*[ 1 1 1 | -5] =
umm idk im not good with these i just started this

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