How do you solve #3u + v + w = 9#, #u + v - w = 5# and #u + 2v + w = 9# using matrices?

1 Answer
Mar 5, 2017

Answer:

#((u),(v),(w)) = ((7/4),(7/2),(1/4))#

Explanation:

In matrix form:

#((3,1,1),(1,1,-1),(1,2,1)) ((u),(v),(w)) = ((9),(5),(9))#

Using Gaussian elimination, we can row reduce, so we set up the augmented matrix:

#((3,1,1),(1,1,-1),(1,2,1)) # # ((9),(5),(9))#

#R2 to R2 - R3 #

#((3,1,1),(0,-1,-2),(1,2,1)) # # ((9),(-4),(9))#

#R3 to R3 -1/3 R1 #

#((3,1,1),(0,-1,-2),(0,5/3,2/3)) # # ((9),(-4),(6))#

#R3 to R3 + 5/3 R2 #

#((3,1,1),(color(red)(0),-1,-2),(color(red)(0),color(red)(0),-8/3)) # # ((9),(-4),(-2/3))#

This is now upper triangular, we are in row echelon form. Now we back-substitute:

From the bottom row:

#-8/3 w = -2/3 #

#implies w = 1/4#

From the middle row:

#- v - 2 w =- 4 #

#implies v = 4 - 2w #

#implies v = 7/2#

From the top row:

#3u + v + w = 9#

#implies u = (9 - v - w)/3#

# = (9 - 7/2 - 1/4)/(3) #

#implies u = 7/4#

So:

#((u),(v),(w)) = ((7/4),(7/2),(1/4))#