Gaussian Elimination

We use Gaussian elimination to solve systems of equations with 3 or more variables Our goal is to get the system reduced so that we can use substitution to solve. We do so by reducing the matrix to this form: 1 0 0 0 1 0 0 0 1 These can be any number Ones across the diagonal and zeros in the left hand bottom corner.

Acceptable row operations: Add two rows together Multiply a row by a non zero number Swap two rows Multiply a row by a number and then add it to another row

Solve Using Gaussian Elimination x+5y+3z = 10 -y+3z = 21 y–2z = 15 Step 1: Write the equations in a matrix of coefficients 1 5 3 10 0 -1 3 21 0 1 -2 15

We now have a The correct form! Step 2: Use row operations to change the matrix to an augmented matrix 1 5 3 10 0 -1 3 21 0 1 -2 15 5 3 10 0 -1 3 21 0 0 1 36 R2+ R3->R3 This -1 needs to be a 1 This 1 needs to be a 0 5 3 10 0 1 -3 -21 0 0 1 36 (-1)R2->R2 We now have a The correct form!

Step 3: Transform the matrix back into equations x+5y+3z = 10 y-3z = -21 z=36 5 3 10 0 1 -3 -21 0 0 1 36

Step 4: Solve using substitution x+5y+3z = 10 y-3z = -21 z=36