Question

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

Answer 1

@UnkleRhaukus

Answer 2

[ 1 1 1 ] [ x ] [-5]
[ 1 -1 3 ] [ y ] = [-1]
[ 4 1 1 ] [ z ] [-2]

Answer 3

yesss ok so what do i do here

Answer 4

[ 1 1 1 | -5]
[ 1 -1 3 |-1]
[ 4 1 1 | -2]

Answer 5

we want make this
[ .........|..]
[ x...... |..]
[ ........ |..]
element into a zero

Answer 6

take away the first row from the second row

Answer 7

how

Answer 8

[ 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)]

Answer 9

ohhh

Answer 10

whats next or is that it

Answer 11

we get

[ 1 1 1 | -5] (R1)
[ 0 -2 2 | -6] (R2)
[ 4 1 1 | -2] (R3)

Answer 12

now we want make this
[ 1.......|..]
[ 0...... |..]
[ x........ |..]
element zero

Answer 13

so take away line row 1 from row 3, four times

Answer 14

soo 111 i think

Answer 15

[ 4 1 1 | -2] - 4*[ 1 1 1 | -5]

=

Answer 16

umm idk im not good with these i just started this

Answer 17

@loser66

Answer 18

@ganeshie8