Ch. 7 – Matrices and Systems of Equations 7.6 – Inverses of Square Matrices

Inverses Identity Matrix (I)= a square matrix with 1s in the main diagonal and 0s everywhere else Two matrices are inverses of each other if they multiply to the identity matrix The inverse of A is denoted A-1 A is said to be invertible if it has an inverse Ex: Prove that B is the inverse of A. Multiply A and B and see if you get the identity matrix! 1 1

Inverses Ex 1: Find the inverse of A. We want to solve A A-1 = I for A-1, but how? Matrix multiplication tells us that we want to solve the following two augmented matrices simultaneously: …so we do that by reducing the adjoined matrix [ A : I ] ! When finding inverses of square matrices, RREF the augmented matrix ! In this problem, we need to RREF the following matrix:

Inverses Ex 1 (cont’d): Find the inverse of A. Add E2 and E1 and replace for E2… Add 6(E2) and E1 and replace for E1…

Inverses Ex 1 (cont’d): Find the inverse of A. So the inverse of A is the right side of the RREF’d matrix. Divide E1 by 2…

Ex 2 : Find the inverse of A. RREF the adjoined matrix! Add -(E2) and E1 and replace for E2… Add -6(E1) and E3 and replace for E3…

Ex 2 (cont’d): RREF the adjoined matrix! Add -(E1) and E2 and replace for E1… Add 4(E2) and E3 and replace for E3…

Ex 2 (cont’d): RREF the adjoined matrix! Add -(E1) and E3 and replace for E1… Add -(E2) and E3 and replace for E2…

Check your solution with your calculator! Ex 2 (cont’d): Check your solution with your calculator! Separate the inverse matrix from the identity to get…

An alternative method to solving systems of equations involves using inverse functions. Ex 3: Use an inverse matrix to solve the system. If we write the system in AX = B form, where A is the coefficient matrix, X is the variable matrix, and B is the constant term matrix, it will look like this: By multiplying A-1 in front of both sides of the equation, we get A to cancel and an equation in X = A-1B form: