How do you solve #x + 3y + z = 3#; #x + 5y + 5z = 1#; #2x + 6y + 3z = 8# using matrices?

Question

2296 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-11T07:02:08

The solution is #((x),(y),(z))=((16),(-5),(2))#

Perform the Gauss-Jordan elimination on the augmented matrix

#A=((1,3,1,|,3),(1,5,5,|,1),(2,6,3,|,8))#

Make the pivot in the first column and the first row

Eliminate the first column, perform the row operations

#R2larrR2-R1#, and #R3larrR3-2R1#

#=((1,3,1,|,3),(0,2,4,|,-2),(0,0,1,|,2))#

Make the pivot in the second column by

#R2larr(R2)/2#

#=((1,3,1,|,3),(0,1,2,|,-1),(0,0,1,|,2))#

Eliminate the second column by

#R1larrR1-3R2#

#=((1,0,-5,|,6),(0,1,2,|,-1),(0,0,1,|,2))#

Make the pivot in the second column, and eliminate the third column

#R1larrR1+5R3# and #R2larrR2-2R3#

#=((1,0,0,|,16),(0,1,0,|,-5),(0,0,1,|,2))#

The solution is

#((x),(y),(z))=((16),(-5),(2))#