How do you solve the system #x+2y-z=6#, #-3x-2y+5z=-12#, and #x-2z=3#?

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Answer 1 · 2017-04-16T19:18:45

Please see the explantion

Write #x-2z=3# as the first row of an Augmented Matrix :

#[ (1,0,-2,|,3) ]#

Add a row for the equation #x+2y-z=6#:

#[ (1,0,-2,|,3), (1,2,-1,|,6) ]#

Add a row for the equation #-3x-2y+5z=-12#:

#[ (1,0,-2,|,3), (1,2,-1,|,6), (-3,-2,5,|,-12) ]#

Perform Elementary Row Operations until an identity matrix is obtained on the left.

We want the coefficient in position #(1,1)# to be 1 and it is, therefore, we do not do a row operation.

We want the other two coefficients in column 1 to be 0, therefore, we perform the following 2 row operations:

#R_2-R_1toR_2#

#[ (1,0,-2,|,3), (0,2,1,|,3), (-3,-2,5,|,-12) ]#

#R_3+3R_1toR_3#

#[ (1,0,-2,|,3), (0,2,1,|,3), (0,-2,-1,|,-3) ]#

Because Row 3 is a scalar multiple of Row 2, one should declare that there is no unique solution.

Let's write the first two rows as linear equations:

#x-2z=3#
#2y+z=3

#z= x/2-3/2#
#y=3/2-z/2

#z= x/2-3/2" [1]"#
#y=9/4-x/4" [2]"#

Equations [1] and [2] will allow you to choose a value of x and then determine the corresponding value of y and z.