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")#