# How do you solve 3 variable equations using augmented matricies

http://www.google.com/search?as_q=&num=10&hl=en&btnG=Google+Search&as_epq=augmented+matrices&as_oq=&as_eq=&lr=&as_ft=i&as_filetype=&as_qdr=all&as_nlo=&as_nhi=&as_occt=title&as_dt=i&as_sitesearch=&as_rights=&safe=active

## To solve a system of three variable equations using augmented matrices, you can follow these steps:

Step 1: Write down the three equations in standard form, where the variables are on the left side and the constants are on the right side.

For example, let's say you have the following three equations:

1) 2x + 3y - z = 4

2) x - 2y + 3z = 1

3) 3x + y + 2z = 6

Step 2: Create the augmented matrix by writing down the coefficients of the variables and the constants in a matrix format. The augmented matrix is an extension of the coefficient matrix by adding a vertical line and writing the constants beside it.

Using the above equations, the augmented matrix would be:

[2 3 -1 | 4]

[1 -2 3 | 1]

[3 1 2 | 6]

Step 3: Use row operations to transform the augmented matrix into its row-echelon form or reduced row-echelon form using techniques such as row swapping, scaling, and elimination.

Performing the row operations would look like this:

1) Multiply Row 1 by 1/2: [1 3/2 -1/2 | 2]

2) Multiply Row 2 by 1/2: [1/2 -1 3/2 | 1/2]

3) Multiply Row 3 by 1/3: [1 1/3 2/3 | 2]

4) Subtract Row 1 from Row 2: [0 -5/2 5/2 | -3/2]

5) Subtract Row 1 from Row 3: [0 -7/6 7/6 | 3/2]

6) Multiply Row 2 by -2/5: [0 1 -1 | 3/5]

7) Multiply Row 3 by -6/7: [0 1 -1 | -9/7]

Step 4: Continue applying row operations until you have the row-echelon form or reduced row-echelon form.

In this example, we have reached the row-echelon form.

[1 3/2 -1/2 | 2]

[0 1 -1 | 3/5]

[0 0 0 | -9/7]

Step 5: Rewrite the row-echelon form as a system of equations. Each row corresponds to an equation.

1) x + (3/2)y - (1/2)z = 2

2) y - z = 3/5

3) 0 = -9/7

Step 6: Solve the system of equations. You may have one or more variables that are dependent or independent.

In this case, we can see that the third equation is inconsistent (0 = -9/7), meaning there is no solution. The system of equations is therefore inconsistent, and you cannot find a unique solution for x, y, and z.

If the third equation had been consistent (0 = 0), this would indicate an infinite number of solutions, and you would need to express the solution in terms of a parameter.

Note: In some cases, you may need to perform additional steps such as back-substitution to find the values of the variables. However, in this particular example, it is not applicable as the system is inconsistent.