How do you solve the system #3x-y+z=5#, #x+3y+3z=-6#, and #x+4y-2z=12#?

1 Answer
Apr 27, 2017

Answer:

#x=3#, #y=1/2#, #z=-7/2#

Explanation:

Given:

#"(i)":color(white)(1/1)3x-y+z=5#
#"(ii)":color(white)(1/1)x+3y+3z=-6#
#"(iii)":color(white)(1/1)x+4y-2z=12#

Note that we can eliminate #z# by adding or subtracting suitable multiples of #"(i)"# to or from #"(ii)"# and #"(iii)"#:

#"(iv)" = "(ii)" - 3"(i)":color(white)(1/1)-8x+6y=-21#
#"(v)" = "(iii)" + 2"(i)":color(white)(1/1)7x+2y=22#

Then we can eliminate #y# by subtracting #3# times #"(v)"# from #"(iv)"#:

#"(vi)" = "(iv)"-3"(v)":color(white)(1/1)-29x = -87#

Dividing both sides of #"(vi)"# by #-29#, we find:

#x = 3#

Substituting this value of #x# in #"(v)"#, we find:

#7(color(blue)(3))+2y = 22#

Hence:

#y = 1/2#

Then from #"(i)"# we find:

#z = 5-3x+y = 5-3(color(blue)(3))+color(blue)(1/2) = -7/2#

#color(white)()#
Matrix formulation

We can represent the original equations as an augmented matrix:

#((3, -1, 1, |, 5), (1, 3, 3, |, -6), (1, 4, -2, |, 12))#

The operations we performed above correspond to various row operations:

Subtract #3xx"row1"# from #"row2"# and add #2xx"row1"# to #"row3"# to get:

#((3, -1, 1, |, 5), (-8, 6, 0, |, -21), (7, 2, 0, |, 22))#

Subtract #3xx"row3"# from #"row2"# to get:

#((3, -1, 1, |, 5), (-29, 0, 0, |, -87), (7, 2, 0, |, 22))#

Divide #"row2"# by #-29# to get:

#((3, -1, 1, |, 5), (1, 0, 0, |, 3), (7, 2, 0, |, 22))#

Subtract #7xx"row2"# from #"row3"# to get:

#((3, -1, 1, |, 5), (1, 0, 0, |, 3), (0, 2, 0, |, 1))#

Divide #"row3"# by #2# to get:

#((3, -1, 1, |, 5), (1, 0, 0, |, 3), (0, 1, 0, |, 1/2))#

Add #"row3"# to #"row1"# and subtract #3xx"row2"# to get:

#((0, 0, 1, |, -7/2), (1, 0, 0, |, 3), (0, 1, 0, |, 1/2))#

Permute the rows, moving #"row1"# to the third row to get:

#((1, 0, 0, |, 3), (0, 1, 0, |, 1/2), (0, 0, 1, |, -7/2))#

Now the left hand #3xx3# submatrix is an identity matrix, we can read off #x#, #y# and #z# from the right hand column.