How to solve the following systems of simultaneous equations using matrices? 1) 2x+3y-z=12 2y+z=7 2y-z=5

1 Answer
Jan 30, 2018

x=2,y=3 and z=1
Shown 1 step at a time. By the time you have checked through 'this lot' the method will be truly fixed in your mind!

Explanation:

By example: using the convention type: R_2 representing row 2

color(brown)("The objective is to have the condition listed below")
color(brown)("All manipulation is to achieve this goal".)

The value of 1 under x in R_1 and zeros under y and z
means that you only have x in that row.

The value of 1 under y in R_2 and zeros under x and z means that you only have y in that row.

The value of 1 under z in R_3 and zeros under x and y means that you only have z in that row.

color(brown)(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~)

color(green)(((x,y,z,|,"answer"),(2,3,-1,|,12),(0,2,1,|,7),(0,2,-1,|,5) ))
color(white)("dddd")R_2xx1/2 Turns the 2 in R_2 into 1
color(white)("dddddd")darr

color(green)(((2,3,-1,|,12),(0,1,1/2,|,7/2),(0,2,-1,|,5) ))
color(white)("ddd"2/2)R_3-2R_2
color(white)("ddddddd")darr

color(green)(((2,3,-1,|,12),(0,1,1/2,|,7/2),(0,0,-2,|,-2) ))
color(white)("ddd"2/2)R_3xx(-1/2)
color(white)("dddddddd")darr

color(green)(((2,3,-1,|,12),(0,1,1/2,|,7/2),(0,0,1,|,1) ))
color(white)("ddd"2/2)R_2-1/2R_3
color(white)("dddddddd")darr

color(green)(((2,3,-1,|,12),(0,1,0,|,3),(0,0,1,|,1) ))
color(white)("ddd"2/2)R_1xx1/2
color(white)("dddddddd")darr

color(green)(((1,3/2,-1/2,|,6),(0,1,0,|,3),(0,0,1,|,1) ))
color(white)("ddd"2/2)R_1-3/2R_2
color(white)("dddddddd")darr

color(green)(((1,0,-1/2,|,3/2),(0,1,0,|,3),(0,0,1,|,1) ))
color(white)("ddd"2/2)R_1+1/2R_3
color(white)("dddddddd")darr

color(green)(((1,0,0,|,2),(0,1,0,|,3),(0,0,1,|,1) ))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check

x=2,y=3 and z=1

2x+3y-z=12 -> 2(2)+3(3)-(1) ->12 color(red)(larr" True")

2y+z=7 -> 2(3)+1 ->7 color(red)(larr" True")

2y-z=5->2(3)-1->5 color(red)(larr" True")