How do you solve #x - y + 4z = 6#, #2x + z = 1# and #x + 5y + z = -9# using matrices?

1 Answer
Mar 25, 2017

Answer:

There are several ways to do it, using matrices. I prefer the use of an augmented matrix

Explanation:

The first equation, #x - y + 4z = 6#, makes the following row in the augmented matrix:

#[ (1,-1,4,|,6) ]#

The second equation, #2x + z = 1#, makes the following row in the augmented matrix:

#[ (1,-1,4,|,6), (2,0,1,|,1) ]#

The third equation, #x + 5y + z = -9#, makes the following row in the augmented matrix:

#[ (1,-1,4,|,6), (2,0,1,|,1), (1,5,1,|,-9) ]#

Now, perform elementary row operations until, you obtain an identity matrix on the left.

#R_2-2R_1toR_2#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (1,5,1,|,-9) ]#

#R_3-R_1toR_3#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (0,6,-3,|,-15) ]#

#R_3-3R_2toR_3#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (0,0,18,|,18) ]#

#R_3/18#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (0,0,1,|,1) ]#

#R_2+7R_3toR_2#

#[ (1,-1,4,|,6), (0,2,0,|,-4), (0,0,1,|,1) ]#

#R_1-4R_2toR_1#

#[ (1,-1,0,|,2), (0,2,0,|,-4), (0,0,1,|,1) ]#

#R_2/2#

#[ (1,-1,0,|,2), (0,1,0,|,-2), (0,0,1,|,1) ]#

#R_1+R_2toR_1#

#[ (1,0,0,|,0), (0,1,0,|,-2), (0,0,1,|,1) ]#

We have an identity matrix on the left, therefore, the solution set is on the right:

#x = 0, y = -2, and z = 1#

Check:

#x - y + 4z = 6#
#2x + z = 1#
#x + 5y + z = -9#

#0 - -2 + 4(1) = 6#
#2(0) + 1 = 1#
#0 + 5(-2) + 1 = -9#

#6 = 6#
#1 = 1#
#-9 = -9#

This checks.