We are given the system

(1)

The matrix associated to this system is

(2)

We want to find the solution set of the system of linear equations represented by this augmented matrix. The first row does not have a leading so multiply the first row with and get the row equivalent matrix

(3)

Next, add the first row and the second row and replace the second row with it to get zero below the leading

(4)

Next, multiply the second row by to get a leading

(5)

Now this matrix is in row-echelon form. We can see that and a back substitution can be used to find from the first row

(6)

We had the option to continue and get a row reduced echelon form that is add first and second row and replace the first row with it

(7)

This provides the solution .

Suppose we are given the matrix

(8)

We want to find the solution set of the system of linear equations represented by this augmented matrix. One option is as followed:

1. Multiply first row with to get a leading

(9)

2. Multiply the second row with and subtract from row one and replace row 2 to get a leading zero

(10)

3. Multiply the second row with

(11)

4. Multiply the third row with and subtract from the first row to get a leading zero

(12)

5. Multiply the third row with to get in the third entry

(13)

Now this matrix is in row echelon form. We see that and could use a back substitution to solve for and However, we continue the reduction process to get a row reduced echelon form

6. Multiply the third row by and subtract the third row from the second row and replace the second row

(14)

7. Add first and third row and replace the first row with it

(15)

Therefore, the solution to this system is .

Suppose we are given the matrix

(16)

We want to find the solution set of the system of linear equations represented by this augmented matrix. One option is as followed:

1. Interchange rows 1 and 2

(17)

2. Multiply the first row by

(18)

3. Multiply the second row by

(19)

4. Multiply the third row by , add to row one and then replace row three

(20)

5. Multiply row 2 by , add to the second row and replace row three

(21)

As we see the matrix is in row echelon form we can use a back substitution to solve the system

(22)

Therefore, the solution is .