How do you solve the system #x+4y-6z=8#, #2x-y+3z=-10#, and #3x-2y+3z=-18#?

2 Answers
Nov 30, 2016

Answer:

The solution is #(x,y,z)=(-4,"  "4,"  "2/3)#.

Explanation:

Write the coefficients in an augmented matrix:
#[(1,4,-6,|,8),(2,-1,3,|,-10),(3,-2,3,|,-18)]#

Our goal is to change this into a matrix that is upper-right triangular; that is, where everything below the diagonal is a zero. We do this by adding linear combinations of the rows together to create new rows (that replace the current ones) and by multiplying rows by constants. Remember: this does not change the information implied by the augmented matrix, it just presents it in a new (and hopefully more beneficial) way.

(Brief notation recap: #R_1#, #R_2#, and #R_3# mean "row 1", "row 2", and "row 3" respectively.)

Let's start with the first cell in row 2. Its value is #2#, which is #2/1=2# times as much as the #1# above it. If we want the #2# to become a #0#, we need to subtract #2R_1 - R_2#:

#2times[(1,4,"-"6,|,8"    ")]#
#ul("    "-[(2,"-"1,3,|,"-"10)])#
#"=    "[(0,9,"-"15,|,26)]#

This new row represents a new valid equation (that is, #0x+9y-15z=26#) which can replace one of the two rows used in forming it. Choosing to keep row 1 the same, we replace row 2:

#[(1,4,-6,|,8),(color(red)0,color(red)(9),color(red)(-15),|,color(red)26),(3,-2,3,|,-18)]#

A similar linear combination is then used to replace row 3—that combination is #3R_1-R_3#, and the following shorthand is used to indicate what row operation has been done to form a new row:

#=>color(white)[(),(),(color(black)(3R_1-R_3))][(1,4,-6,|,8),(0,9,-15,|,26),(color(red)0,color(red)14,color(red)(-21),|,color(red)42)]#

All the coefficients in #R_3# are multiples of 7, so let's reduce:

#=>color(white)[(),(),(color(black)(1/7*R_3))][(1,4,-6,|,8),(0,9,-15,|,26),(0,2,-3,|,6)]#

#=>color(white)[(),(),(color(black)(9R_3-2R_2))][(1,4,-6,|,8),(0,9,-15,|,26),(0,0,3,|,2)]#

Our newest #R_3# says #3z=2#, which means that #z=2/3#. And since the equation directly above #R_3# only involves one additional unknown (that is, #y#), we can plug this solved value for #z# into it to solve for #y#:

#9y-15z=26#
#=>9y-15(2/3)=26#
#=>9y-10=26#
#=>9y=36#
#=>y=4#

With the matrix in the upper-triangular form, solving for variables is like falling dominoes. Knowing #z# gives us #y#, and knowing #z# & #y# gives us #x#:

#x+4y-6z=8#
#=>x+4(4)-6(2/3)=8#
#=>x+16-4=8#
#=>x="-"4#

Our final solution is #(x,y,z)=(-4,"  "4,"  "2/3)#.
(It is best to check your solution to make sure it works; I will save that as an exercise.)

Note:

Many teachers will encourage you to reduce each new row you create so that the leading coefficients are #1#. This is perfectly fine, and you'll get the same solutions. It just means your matrix will likely have fractions in it. I chose not to reduce my rows for the sake of making the matrices easier to read.

Nov 30, 2016

Answer:

Here is an alternate approach that turns the coefficient matrix into the identity matrix, thus making the augmented column the solution.

Explanation:

#"                         "[(1,4,-6,|,8),(2,-1,3,|,-10),(3,-2,3,|,-18)]#

#=>color(white)[(),(color(black)(2R_1-R_2)),()][(1,4,-6,|,8),(0,9,-15,|,26),(3,-2,3,|,-18)]#

#=>"      "color(white)[(),(color(black)(1/9*R_2)),()][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(3,-2,3,|,-18)]#

#=>color(white)[(),(),(color(black)(3R_1-R_3))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,14,-21,|,42)]#

#=>"      "color(white)[(),(),(color(black)(1/7*R_3))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,2,-3,|,6)]#

#=>color(white)[(),(),(color(black)(R_3-2R_2))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,0,1/3,|,2/9)]#

#=>"       "color(white)[(),(),(color(black)(3*R_3))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,0,1,|,2/3)]#

#=>color(white)[(),(color(black)(R_2+5/3R_1)),(3*R_3)][(1,4,-6,|,8),(0,1,0,|,4),(0,0,1,|,2/3)]#

#=>color(white)[(color(black)(R_1-4R_2)),(),(3*R_3)][(1,0,-6,|,-8),(0,1,0,|,4),(0,0,1,|,2/3)]#

#=>color(white)[(color(black)(R_1+6R_3)),(),(3*R_3)][(1,0,0,|,-4),(0,1,0,|,4),(0,0,1,|,2/3)]#

These equations now say

#x"            "="-"4#
#"      "y"      "=4#
#"            "z=2//3#

So our answer is as before.