Question

How to fi the final form of the augmented matrix when solving for the inverse of this system of equations? `2x+3y=4`nd `4x+5y=5`

A) `[(1,0),(0,1)]|[(5/2,-3/2),(-3,4)]`

B) `[(1,0),(0,1)]|[(-5/2,3/2),(2,-1)]`

C) `[(1,0),(0,1)]|[(5,-3),(-4,2)]`

D) `[(1,0),(0,1)]|[(5/2,-3/2),(-2,1)]`

I have zero clue what the question wants me to do... I was never taught that you could find the inverse of an augmented matrix.... Plz halp :/

Answer 1

`color(blue)(B)`

Explanation:

The question is asking you to identify the inverse of the matrix formed with the coefficients of the linear system. We could find the inverse of this matrix in different ways. Looking at the choices would imply they want you to find the inverse by forming an augmented matrix with the `2xx2` identity matrix.

We will denote the matrix of the coefficients as `bbA`

We have:

`2x+3y=4` and `4x+5y=5`

So matrix `bbA` is:

`bbA=[(2,3),(4,5)]`

And the `2xx2` identity matrix is:

`bbI=[(1,0),(0,1)]`

Augmenting:

`[(2,3),(4,5)]|[(1,0),(0,1)]`

Now our task is to perform row operations on this rectangular matrix, with the objective of transforming the matrix on the left hand side into the identity matrix and as this is formed the right hand matrix becomes the inverse of `bbA`. I will align these as a single matrix from here on, because it's easier to format.

`[(2,3,1,0),(4,5,0,1)]`

The notation for the row operations will be:

`R1=R1+2*R2`

This reads, row 1 is row 1 plus 2 times row 2.

`[(2,3,1,0),(4,5,0,1)]`

`R2=R2-2*R1`

`[(2,3,1,0),(0,-1,-2,1)]`

`R1=R1+3*R2`

`[(2,0,-5,3),(0,-1,-2,1)]`

`R1=R1*1/2`

`[(1,0,-5/2,3/2),(0,-1,-2,1)]`

`R2=R2*-1`

`[(1,0,-5/2,3/2),(0,1,2,-1)]`

We have now found the inverse of `bbA`

i.e.

`bb(A^-1)=[(-5/2,3/2),(2,-1)]`

This is choice B.