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

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.