Systems of simultaneous linear equations are solved mechanically through the use of a method called Gauss elimination. This method uses a matrix formed by the constant coefficients in the equations augmented by the vector formed by the equation solutions. A series of multiplication-subtraction operations are performed to create a triangular matrix, and then new values from the matrix are substituted back into the equations to determine the values for the variables. The matrix should have the same number of rows as there are variables in the problem. Otherwise, there will be no unique solution.
Simultaneous Linear Equations
Write your equations in standard form. Create the augmented matrix from the coefficients and equation solutions:
x + y + z = 6
x + 2y + 2z = 11
2x + 3y - 4z = 3
Multiply the first row by a constant factor and subtract those values from the second row. Choose a factor that will leave a zero in the first position of the second row after the subtraction. Repeat for the third row. In this case, the factor for the operation on the second row is 1, and the factor for the operation on the third row operation is 2.
Multiply the second row by a factor that will set the second term equal to 1. In this case, the factor is -1.
Multiply the second row by a factor and subtract those values from the third row as before. For this example, the factor is -1.
Multiply the third row by a factor that will set the third term equal to 1. In this example, the third row is (0, 0, -7, -14) after the row operations, so a factor of -1/7 should be used. This completes the "forward elimination" portion of the problem.
Rewrite the equations using the new coefficients and solutions:
x + y + z = 6
0x + y + z = 5
0x + 0y + z = 2
Substitute the known values back into the equations to determine the values of x, y and z. This is called "back substitution":
0x + 0y + z = 2; z = 2
0x + y + 2 = 5; y = 3
x + 3 + 2 = 6; x = 1
- Linear Algebra and Its Applications, 3rd ed.; Gilbert Strang; 1988