Matrix Inversion Method

Matrix Inversion Method

This method can be applied only when the coefficient matrix is a square matrix and non-singular.

Consider the matrix equation

     AX   =  B ,      â€¦ (1)

where A is a square matrix and non-singular. Since A is non-singular, A^−1 exists and A^−1 A = AA^−1 = I. Pre-multiplying both sides of (1) by A^−1, we get A^−1 ( AX ) = A^−1B. That is, ( A^−1 A) X = A^−1B. Hence, we get X = A^−1B.

Example 1.22

Solve the following system of linear equations, using matrix inversion method:

5x + 2 y = 3, 3x + 2 y = 5 .

Solution

The matrix form of the system is AX = B , where 

We find |A| =   = 10 - 6= 4 ≠ 0. So, A^−1  exists and A^−1 = 

Then, applying the formula X = A^−1B , we get

So the solution is (x = −1, y = 4).

Example 1.23

Solve the following system of equations, using matrix inversion method:

2x₁ + 3x₂ + 3x₃ = 5,

x₁ – 2x₂ + x₃ = -4,

3x₁ – x₂ – 2x₃ = 3

Solution

The matrix form of the system is AX = B,where

So, the solution is ( x₁ = 1, x₂ = 2, x₃ = −1) .

Example 1.24

If , find the products AB and BA and hence solve the system of equations x − y + z = 4, x – 2y – 2z = 9, 2x + y +3z =1.

Solution

Writing the given system of equations in matrix form, we get

Hence, the solution is (x = 3, y = - 2,  z = −1).