Question

How do you set up and solve the following system using augmented matrices `x-2y=7, 3x+4y=1`?

Answer 1

Please see the explanation.

Explanation:

The equation `x-2y=7` gives us the following line in the augmented matrix:

#[ (1,-2,|,7) ]#

The equation `3x+4y=1` adds the second line to the augmented matrix:

#[ (1,-2,|,7), (3,4,|,1) ]#

Perform elementary row operation until an identity matrix is obtained on the left, then the column on the right will contain the solutions.

We want the coefficient in position `(1,1)` to be 1 and it is, therefore, no operation is needed.

We want the other coefficient in column 1 to be zero, therefore, we perform the following row operation:

`R_2-3R_1toR_2`

#[ (1,-2,|,7), (0,10,|,-20) ]#

We want the coefficient is position `(2,2)` to be one, therefore, we perform the following row operation:

`R_2/10toR_2`

#[ (1,-2,|,7), (0,1,|,-2) ]#

We want the other coefficient in column to be 0, therefore, we perform the following row operation:

`R_1+2R_2toR_1`

#[ (1,0,|,3), (0,1,|,-2) ]#

We have an identity matrix on the left, therefore the column on the right contains the solution: `x = 3 and y=-2`

Check:

`x-2y=7`
`3x+4y=1`

`3-2(-2)=7`
`3(3)+4(-2)=1`

`7=7`
`1=1`

This checks.