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

2 Answers
Dec 29, 2016

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

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))