How do you solve #[(2x+y),(x-3y)]=[(5), (13)]#?

2 Answers
Dec 29, 2016

Answer:

Create and augmented matrix and the perform row operations until you obtain an identity matrix on the left of the divide.

Explanation:

The matrix of coefficients is:

#A = [ (2,1), (1,-3) ]#

The matrix of constant terms is:

#B = [ (5), (13) ]#

The method that I use is to make an Augmented Matrix

#(A|B) = [ (2,1,|,5), (1,-3,|,13) ]#

And then perform Elementary Row Operations until I obtain an Identity Matrix on the left side of the divide.

#R_2 harr R_1#

#[ (1,-3,|,13), (2,1,|,5) ]#

#R_2 - 2R_1 to R_2#

#[ (1,-3,|,13), (0,7,|,-21) ]#

#R_2/7 to R_2#

#[ (1,-3,|,13), (0,1,|,-3) ]#

#R_1 + 3R_2 to R_1#

#[ (1,0,|,4), (0,1,|,-3) ]#

We have obtained an identity matrix.

The values of the unknowns can be read on the right:

#x = 4 and y = -3#

Check:

#2(4) + -3 = 5#
#4 - 3(-3) = 13#

#5 = 5#
#13 = 13#

This checks

Dec 29, 2016

Answer:

The answer is #((x),(y))=((4),(-3))#

Explanation:

We rewrite the equation in matrix form

#((2,1),(1,-3))((x),(y))=((5),(13))#

#((x),(y))=((2,1),(1,-3))^(-1)((5),(13))#

Let #A=((2,1),(1,-3))#

We calculate

#detA=| (2,1), (1,-3) |=-6-1=-7#

#detA!=0#, therefore the matrix is invertible

We start by calculating

the matrix of cofactor

#C=((-3,-1),(-1,2))#

Then, we calculate the transpose

#C^T=((-3,-1),(-1,2))#

The inverse is

#A^(-1)=1/-7((-3,-1),(-1,2))#

#=((3/7,1/7),(1/7,-2/7))#

Verification

#A*A^(-1)=((2,1),(1,-3))*((3/7,1/7),(1/7,-2/7))=((1,0),(0,1))#

Therefore,

#((x),(y))=((3/7,1/7),(1/7,-2/7))((5),(13))#

#((x),(y))=((4),(-3))#