How do you solve #-5x+6y-z= -26# and #x-4y-4z= -27# and #6x+y+z = 38# using matrices?

1 Answer
Jun 28, 2016

Answer:

#[(1+0+0 |+5),(0+1+0|+1),(0+0+1|+7)]#

#x=5; y=1; z=7#

Explanation:

AS a check

Tony B
#"Target values"->x=5; y=1; z=7#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("These can go wrong very easily. Particularly if fractions are involved.")#

#" "x" "y" "z" " =#
#[(-5+6-1 |-26),(1-4-4|-27),(6+1+1|38)]#
#Row1 -:(-5)" and "Row3-:6#
#color(red)(" "darr)#

#[(1-6/5+1/5 |+26/5),(1-4-4|-27),(1+1/6+1/6|19/3)]#
#Row2-Row1 #
#color(red)(" "darr)#

#[(1-6/5+1/5 |+26/5),(0-14/5-21/5|-161/5),(1+1/6+1/6|19/3)]#
#Row2xx(-5/14)#
#color(red)(" "darr)#

#[(1-6/5+1/5 |+26/5),(0+1+3/2|+23/2),(1+1/6+1/6|19/3)]#
#Row1+6/5Row2#
#color(red)(" "darr)#

#[(1+0+2 |+19),(0+1+3/2|+23/2),(1+1/6+1/6|19/3)]#
#Row3-Row1#
#color(red)(" "darr)#

#[(1+0+2 |+19),(0+1+3/2|+23/2),(0+1/6-11/6|-38/3)]#
#Row3xx6#
#color(red)(" "darr)#

#[(1+0+2 |+19),(0+1+3/2|+23/2),(0+1-11|-76)]#
#Row3-Row2#
#color(red)(" "darr)#

#[(1+0+2 |+19),(0+1+3/2|+23/2),(0+0-25/2|-175/2)]#
#Row3xx(-2/25)#
#color(red)(" "darr)#

#[(1+0+2 |+19),(0+1+3/2|+23/2),(0+0+1|+7)]#
#Row1-2Row3" and "Row2-3/2Row3#
#color(red)(" "darr)#

#[(1+0+0 |+5),(0+1+0|+1),(0+0+1|+7)]#