How do you solve #((4, 8), (2, 5))((x), (y))=((0),(6))#?

2 Answers
Feb 12, 2016

Answer:

#((x),(y))=((-12),(6))#

Explanation:

To solve #((4,8),(2,5))((x),(y))=((0),(6))#, we need to first find inverse of the matrix #((4,8),(2,5))#.

The general form for inverse of matris #((a,b),(c,d))# is #1/(ad-bc)((d,-b),(-c,a))#.

Hence inverse of #((4,8),(2,5))# is #1/(4*5-8*2)((5,-8),(-2,4))# or #1/4((5,-8),(-2,4))#. Multiplying both sides of #((4,8),(2,5))((x),(y))=((0),(6))# by this gives

#((x),(y))=1/4((5,-8),(-2,4))((0),(6))# .............(A)

as #1/4((5,-8),(-2,4))((4,8),(2,5))=((1,0),(0,1))#

Solving (A) by simple matrix multiplication gives us

#((x),(y))=1/4((-48),(24))=((-12),(6))#

Feb 27, 2016

Answer:

#(x,y)=(-12,6)#

Explanation:

(this is just an alternative method to that provided by Shwetank Mauria that doesn't require the use of an inverse matrix)

#((4,8),(2,5))xx((x),(y)) = ((4x+8y),(2x+5y))# (by standard matrix multiplication)

Therefore
#color(white)("XXX")((4,8),(2,5))xx((x),(y))= ((0),(6))#
is equivalent to
[1]#color(white)("XXX")4x+8y=0#
[2]#color(white)("XXX")2x+5y=6#

If we multiply equation [2] by #2# and subtract the result form equation [1], we get
[3]#color(white)("XXX")-2y=-12#
and from this
[4]#color(white)("XXX")y=6#

Substituting #6# for #y# back in equation [1] gives
[5]#color(white)("XXX")cancel(4)x+cancel(8)^2xx(6)=0#
and
[6]#color(white)("XXX")x=-12#