The Inverse of a Square Matrix

Introduction:

If a is a real number then its multiplicative inverse is = a^-1 and ax = 1

Definition: The Inverse of a Square Matrix:

Let A be an n x n matrix and I_n be the n x n identity matrix. If there exists matrix A^-1 such that

A A^-1 = A^-1 A =I_n

then  A^-1  is called the inverse of A. Read A^-1 as “ A inverse”.

Example: Check whether B is the Inverse of A or not.

A = and B =

AB =   =  = I₂

BA = = = I₂

AB = BA = I₂

Hence B is the inverse of A.

Definition: if a matrix A has an inverse, A is called invertible (or nonsingular). Otherwise, A is called singular.

If A is a 2 x 2 matrix given by A = , then A is invertible only if ad-bc ≠ 0.

The inverse of the matrix is unique.

Formula to find the inverse of the matrix:

A^-1 = (adj A) , with |A|≠ 0 i.e. A is non singular.

Here adj A = , where A_ij is the cofactor of a_ij

Properties:

• 1. If A and B are any two non-singular matrices if the same order, then AB is also non-singular and

   (AB)^-1 =B^-1A^-1

  i.e. the inverse of a product is the product of the inverse taken in the reverse order.

• 2. If A and B are matrices conformable to multiplication, then (AB)^T = B^T A^T
• 3. For any non-singular matrix A, (A^T)^-1 = (A^-1)^T

Solution of a system of linear equations by matrix inversion method:

Let A be a non-singular matrix.

Consider a system of n linear non-homogeneous equations in n unknow1ns x_1, x₂,x₃,…,x_n.

a₁₁x₁+a₁₂x₂+…………….+a_1nx_n=b₁.

a₂₁x₁+a₂₂x₂+…………….+a_2nx_n=b₂.

.

.

.

a_n1x₁+a_n2x₂+…………….+a_nnx_n=b_n.

This is of the form AX=B

where A =

If A is non-singular, then A^-1 exists.

In the equation AX=B, pre multiply by A^-1, we get

A^-1 (AX) =A^-1B

(A^-1A)X=A^-1B ,(by associative property)

IX=A^-1B , by inverse property)

X=A^-1B     is the solution of the equation AX=B.

Example:

Solve by matrix inversion method x + y = 3, 2x + 3y = 8

The given equation can be written in the form of

Which is of the form AX=B

Here |A| = = 1 ≠ 0

Since A is non-singular, A^-1 exists.

A^-1 =

The solution is X = A^-1B =

Solution: x=1, y=2