. . (iii) Gaussian Elimination Method This method can be applied even if the coefficient matrix is singular matrix and rectangular matrix. It is essentially the method of substitution which we have already seen.
In this method, we transform the augmented matrix of the system of linear equations into row-echelon form and then by back-substitution, we get the solution. Example . Solve the following system of linear equations, by Gaussian elimination method : Transforming the augmented matrix to echelon form, we get ↔ ÷ − ÷ − ) , The equivalent system is written by using the echelon form: = , … ( ) = , … ( ) z = . … ( ) From ( ), we get z = .
Substituting z = in ( ), we get y = = − . Substituting z in ( ), we get x = −× − −× So, the solution is ( . Note. The above method of going from the last equation to the first equation is called the method of back substitution.
Example . The upward speed v t ( )of a rocket at time t is approximated by v t at bt ( ) £ £ where a b , , and are constants. It has been found that the speed at times t , and t = seconds are respectively, , , and miles per second respectively. Find the speed at time t = seconds.
(Use Gaussian elimination method.) - - Applications of Matrices and Determinants Since v v v ( ) ( ) ( ) , we get the following system of linear equations = , = , + = . We solve the above system of linear equations by Gaussian elimination method. Reducing the augmented matrix to an equivalent row-echelon form by using elementary row operations, we get ] A B R R ÷ − ), ÷ − →− Writing the equivalent equations from the row-echelon matrix, we get By back substitution, we get c −= So, we get v t ( ) Hence, v ( + = + =