How do you solve #1/2x+y=5# and #-x-2y=4# using matrices?

1 Answer
Alan P.
Feb 25, 2016

There are no solutions; the equations represent two parallel lines which do not intersect.

Explanation:

If you multiply the first equation by #(-2)# you will see that the new version of the first equation and the second equation have identical expressions on the left side but different values on the right side.

I you try to handle this as a matrix problem:

#((1/2,1,5),(-1,-2,4))#

using Cramer's Rule (determinants)
#x=|D_x|/|D|color(white)("XXX")y=|D_y|/|D|#

#|D| = |(1/2,1),(-1,-2)| = 1/2xx(-2)-(-1)xx2=0#

but division by #0# is not valid, so #x# and #y# can not be evaluated.

using Gauss-Jorden Method
#((1/2,1,5),(-1,-2,4))#

#rarr ((1,2,10),(-1,-2,-4))#

#rarr ((1,2,10),(0,0,6))#

both the #x# and #y# columns are zero, so the second row of this matrix implies:
#color(white)("XXX")0*x+0*y=6#
which is clearly impossible.

Related questions
See all questions in Solving a System of Equations Using a Matrix
Impact of this question
1410 views around the world
You can reuse this answer
Creative Commons License