The system of equation can be seen as an equation of matrices:
[A] [X]=[K]
For that system:
we have:
If exists [A]-1 exists,
[A].[X] = [K]
[A]-1.[A].[X] = [A]-1.[K]
[I]. [X] = [A]-1.[K]
[X] = [A]-1.[K]
Hence, to solve the
equation, we have to calculate [A]-1 and [A]-1.[K]
(Please read the chapter about the matrix to learn more about the inverse of a matrix or check out the example below).
Note:
The conditions for the existence of the inverse of the coefficient matrix are:
1. The system must have the same number of equations as variables, that is, the coefficient matrix of the system must be square.
2. The determinant of the coefficient matrix must be non-zero.
Example
To use this method follow the steps demonstrated on the following system:
Step 1: Rewrite the system using matrix multiplication:
and writing the coefficient matrix as A, we have
.
Step 2: FInd the inverse of the coefficient matrix A. In this case the inverse is
Step 3: Multiply both sides of the equation (that you wrote in step #1) by the matrix A-1.
On the left you'll get
.
On the right, you get
and so the solution is