Recall that a simple linear equation of the form ax = b has the solution x = b/a = a^-1.b, a ≠ 0.

As noted in earlier modules, a system of linear equations can be written as AX = B, where A and B are known matrices and X is the variable matrix. Note that this equation resembles a simple linear equation. The question is does the solution look like X = A^-1. B? If so, what is A^-1 and how to find it?

In this module we will learn about matrix inverse and use the same, if it exists, to find the solution to a system of linear equations.

Review: Augmented matrices and Gauss-Jordan Elimination method.

Learning Objectives:

• Square matrix, singular and non-singular matrices
• Identify matrix
• Inverse of a non-singular matrix
• Use the inverse, when it exists, to find the solution to a system of linear equations.