Question

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

Answer 1

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

Answer 2

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))`