Invertible Matrices

Last Updated : 12 Mar, 2021

A matrix is a representation of elements, in the form of a rectangular array. A matrix consists of rows and columns. Horizontal lines are known as rows and vertical lines are known as columns. The order of a matrix is defined as number of rows × number of columns. If the number of rows in a matrix is “m” and the number of columns is “n” then the order of the matrix is represented as “m×n”.

The inverse of a Matrix

Suppose ‘A’ is a square matrix, now this ‘A’ matrix is known as invertible only in one condition if their another matrix ‘B’ of the same dimension exists, such that, AB = BA = I_n where I_nis known as identity matrix of the same order and matrix ‘B’ is known as the inverse of the matrix ‘A’. The inverse of a matrix can be represented as A^-1. It is also known as non-singular matrix or nondegenerate matrix.

For example:

A = $\begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix}$
and B = $\begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix}$
On multiplying A and B you get,
AB = $\begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix} \begin{bmatrix} 5&-6 \\ -4&5 \end{bmatrix}$
AB = $\begin{bmatrix} 25-24 &-30+30 \\ 20-20&-24+25 \end{bmatrix}$
AB = $\begin{bmatrix}1 &0 \\ 0&1 \end{bmatrix}$
AB = I ………. (1)
Similarly, you can get BA by multiplying matrix B and matrix A.
BA = $\begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix} \begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix}$
BA = $\begin{bmatrix} 25-24 &30-30 \\ -20+20&-24+25 \end{bmatrix}$
BA = $\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$
BA = I………… (2)
From (1) and (2), you can see that AB = BA = I_n
Hence, A is an invertible matrix and inverse of matrix A is matrix B. This can be written as A^-1 = B.
If B is inverse matrix for A then also, A is inverse matrix for B. So, you can write B^-1 = A.

Note: The necessary and sufficient condition for a square matrix A to possess the inverse is that the matrix should not be singular. A matrix is called a singular matrix, if the determinant of the matrix is zero i.e. |A| = 0. So |A| ≠ 0 for a matrix A which is invertible.

Application of invertible matrix:

The application of invertible matrix is:

Least-squares or Regression
Simulations
MIMO Wireless Communications

Matrix inversion methods:

Using the following methods you can find the other matrix say ‘B’ matrix which is the inverse of matrix ‘A’:

Gaussian Elimination
Newton’s Method
Cayley-Hamilton Method
Eigen Decomposition Method

Example: Check whether matrix A= $\begin{bmatrix} 0 &1 &2 \\ 1&2 &3 \\ 3& 1 & 1 \end{bmatrix}$ is invertible or not. And if A is invertible then check whether matrix B = $\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix}$ is inverse of matrix A or not.

Solution:

First we check whether matrix A is invertible or not.
|A| = 0×(2-3) – 1×(1-2) + 3×(3-4)
|A| = 0+1-3
|A| = -2 ≠0
Hence. matrix A is invertible.
Now, check whether AB=BA=I_n or not
AB = $\begin{bmatrix} 0 &1 &2 \\ 1&2 &3 \\ 3& 1 & 1 \end{bmatrix}\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix}$
AB = $\frac{-1}{2} \begin{bmatrix} 0 &1 &2\\ 1&2 &3 \\ 3& 1& 1 \end{bmatrix} \begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix}$
AB = $\frac{-1}{2}\begin{bmatrix} 0+8-10 &0-6+6 &0+2-2 \\ -1+16-15&1-12+9 &-1+4-3 \\ -3+8-5& 3-6+3 & -3+2-1 \end{bmatrix}$
AB = $\frac{-1}{2} \begin{bmatrix} -2&0 &0\\ 0&-2 &0 \\ 0& 0& -2 \end{bmatrix}$
AB = $\begin{bmatrix} 1&0 &0\\ 0&1 &0 \\ 0& 0& 1 \end{bmatrix}$
AB = I
BA = $\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix} \begin{bmatrix} 0 &1 &2 \\ 1&2 &3 \\ 3& 1 & 1 \end{bmatrix}$
BA = $\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix} \begin{bmatrix} 0 &1 &2\\ 1&2 &3 \\ 3& 1& 1 \end{bmatrix}$
BA = $\frac{-1}{2}\begin{bmatrix} 0+1-3 &-1+2-1 &-2+3-1 \\ 0-6+6&8-12+2 &16-18+2 \\ 0+3-3& -5+6-1 & -10+9-1 \end{bmatrix}$
BA = $\frac{-1}{2} \begin{bmatrix} -2&0 &0\\ 0&-2 &0 \\ 0& 0& -2 \end{bmatrix}$
BA = $\begin{bmatrix} 1&0 &0\\ 0&1 &0 \\ 0& 0& 1 \end{bmatrix}$
BA = I
You can see, AB = BA = I
Hence, matrix B is inverse of matrix A.

Theorems of Invertible Matrix

Theorem 1: Every invertible matrix posses a unique inverse.

Proof:

Let ‘A’ be an n×n invertible matrix.
Let us considered B and C be two inverse of A.
Then, AB = BA = I …….(1)
and AC = CA = I …….. (2)
From (1) you have
C(AB) = C(I_n) = C ……..(3)
From (2) you have
(CA)B = I_n(B) = B ……. (4)
Since, C(AB) = (CA)B [Associativity Law]
So, C = B
Hence, it is proved.

Example:

Let A = $\begin{bmatrix} 2 &9 \\ 1& 7 \end{bmatrix}$ , B = $\begin{bmatrix} b_{1} &b_{2} \\ b_{3}& b_{4} \end{bmatrix}$ , and C = $\begin{bmatrix} c_{1} &c_{2} \\ c_{3}& c_{4} \end{bmatrix}$
If both matrices B and C are inverse of matrix A then,
AB = BA = I and
AC = CA = I
Taking AB = I
$\begin{bmatrix} 2 &9 \\ 1& 7 \end{bmatrix} \begin{bmatrix} b_{1} &b_{2} \\ b_{3}& b_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$
$\begin{bmatrix} 2b_{1}+9b_{3} &2b_{2}+9b_{4} \\ b_{1}+7b_{3}& b_{2}+7b_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$
From above, you get 4 equations
2 b1 + 9 b3 = 1 ………. (1)
b1 + 7 b3 = 0 …………….(2)
2 b2 + 9 b4 = 0 …………..(3)
b2 + 7 b4 = 1 ……………….(4)
after solving these 4 equations, you will get
b1 = 7/5
b2 = -9/5
b3 = -1/5
b4 = 2/5
So matrix B = $\begin{bmatrix} \frac{7}{5} &\frac{-9}{5} \\ \frac{-1}{5}& \frac{2}{5} \end{bmatrix}$ ………… (5)
Now, consider AC = I
$\begin{bmatrix} 2 &9 \\ 1& 7 \end{bmatrix} \begin{bmatrix} c_{1} &c_{2} \\ c_{3}& c_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$
$\begin{bmatrix} 2c_{1}+9c_{3} &2c_{2}+9c_{4} \\ c_{1}+7c_{3}& c_{2}+7c_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$
From above, you get 4 equations
2 b1 + 9 b3 = 1 ………. (6)
b1 + 7 b3 = 0 …………….(7)
2 b2 + 9 b4 = 0 …………..(8)
b2 + 7 b4 = 1 ……………….(9)
After solving these 4 equations, you will get
b1 = 7/5
b2 = -9/5
b3 = -1/5
b4 = 2/5
So matrix B = $\begin{bmatrix} \frac{7}{5} &\frac{-9}{5} \\ \frac{-1}{5}& \frac{2}{5} \end{bmatrix}$ ………..(10)
From (9) and (10) you can see that matrix B and C are equal.
Hence, it is proved that any invertible matrix posses unique inverse.

Theorem 2 : If A, B be two n-rowed non-singular matrices then AB is also non-singular and (AB)^-1= B^-1A^-1

Proof:

|A| ≠0, |B| ≠0
So, |AB| ≠0
Let a matrix C = B^-1A^-1
(AB)C = (AB)B^-1A^-1
(AB)C = A(BB^-1)A^-1
(AB)C = AI_nA^-1
(AB)C = AA^-1
(AB)C = I_n
C(AB) = B^-1A^-1(AB)
C(AB) = B^-1A^-1AB
C(AB) = B^-1B
C(AB) = I_n
Since (AB)C = C(AB) = I_n
Hence, C is inverse of (AB)
So (AB)^-1 = B^-1A^-1

Example:

Let A = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$ and B = $\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$
AB = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix} \begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$
AB = $\begin{bmatrix} 3+4 &0+8 \\ 6+0& 0 \end{bmatrix}$
AB = $\begin{bmatrix} 7 &8 \\ 6& 0 \end{bmatrix}$
(AB)^-1 = \frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix} ………. (1)
Inverse of a matrix can be obtained by the given formula
A^-1 = adjoint of matrix A/ |A|
B = $\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$
B^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix}$
A = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$
A^-1 = $\frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$
B^-1A^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix} \frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$
B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8+0 \\ 0-6& 4+3 \end{bmatrix}$
B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}$ …….. (2)
From (1) and (2), you can see that (AB)^-1 = B^-1A^-1
Hence, it is proved that (AB)^-1 = B^-1A^-1

Properties of inverse of a square matrix

1. (A^-1)^-1= A

Proof:

If A is an invertible matrix then
AA^-1 = I
Taking inverse on both sides
(AA^-1)^-1 = I^-1
(A^-1)^-1A^-1 = I [from theorem 2 (AB)^-1 = B^-1A^-1]
Multiplying by A on both sides
(A^-1)^-1A^-1A = IA
(A^-1)^-1I = A
(A^-1)^-1 = A
Hence, it is proved that (A^-1)^-1 = A

Example:

Let A = $\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$
|A| = 12 – 2 = 10
adj A = $\begin{bmatrix} 4 &-1 \\ -2 & 3 \end{bmatrix}$
A^-1 = adj A /|A|
A^-1 = $\frac{1}{10}\begin{bmatrix} 4 &-1 \\ -2 & 3 \end{bmatrix}$
(A^-1)^-1 = adj (A^-1) / |A^-1|
adj(A^-1) = $\frac{1}{10}\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$
|A^-1| = (12 – 10)/100
|A^-1| = 1/10
(A^-1)^-1 = $10*\frac{1}{10}\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$
(A^-1)^-1 = $\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$
From above you can see that (A^-1)^-1 = A

2. (A₁A₂A₃………..A_n)^-1 = A_n^-1A_n-1^-1……….A₂^-1A₁^-1

You can also write it:

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

(ABC)^-1 = A^-1B^-1C^-1

Proof:

This can be proved by mathematical induction
for n = 2
(A₁A₂)^-1 = A₂^-1A₁^-1 ……….(1)
This statement is true. [by theorem 2]
Let this is true for n = k
(A₁A₂A₃……….A_k)^-1 = A_k^-1…………A₂^-1A₁^-1……..(2)
For n = k+1, you have to prove this.
(A₁A₂A₃……….A_kA_k+1)^-1
=((A₁A₂A₃………A_k)A_k+1)^-1
=((A_k^-1…………A₂^-1A₁^-1)A_k+1)^-1
=(A_k+1)^-1 (A_k^-1…………A₂^-1A₁^-1) [using theorem 2]
= A_k+1^-1A_k^-1………….A₂^-1A₁^-1
Hence, it is proved.

Example:

Suppose there are two matrices A and B,
Let A = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$
and B = $\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$
AB = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix} \begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$
AB = $\begin{bmatrix} 3+4 &0+8 \\ 6+0& 0 \end{bmatrix}$
AB = $\begin{bmatrix} 7 &8 \\ 6& 0 \end{bmatrix}$
(AB)^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}$ ………. (1)
Inverse of a matrix can be obtained by the given formula
A^-1 = adjoint of matrix A/ |A|
B = $\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$
B^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix}$
A = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$
A^-1 = $\frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$
B^-1A^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix} \frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$
B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8+0 \\ 0-6& 4+3 \end{bmatrix}$
B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}$ …….. (2)
From (1) and (2), you can see that (AB)^-1 = B^-1A^-1
Hence, it is proved that (AB)^-1 = B^-1A^-1

3. AA^-1= A^-1A = I_n

Proof:

A matrix is invertible if AA^-1 = I
Multiply by A on both sides
AAA^-1 = AI
AI = A
Multiplying by A^-1 on both sides
A^-1AI = A^-1A
I = A^-1A
Hence, it is proved that AA^-1 = I = A^-1A

Example:

Let A = $\begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}$
|A| = 3 + 2 = 5
adj A = $\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix}$
A^-1= adj A \|A|
A^-1 = $\frac{1}{5}\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix}$
Now to prove AA^-1 = A^-1A = I
Taking left hand side
AA^-1
$\begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}\frac{1}{5}\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix}$
$\frac{1}{5}\begin{bmatrix} 3 +2 &-2 +2\\ -3+3 &2+3 \end{bmatrix}$
$\frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix}$
$\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}$
The above matrix is equal to the identity matrix.
Now taking right side
A^-1A
$\frac{1}{5} \begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix} \begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}$
$\frac{1}{5}\begin{bmatrix} 3+2 & 6-6\\ 1-1&2+3 \end{bmatrix}$
$\frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix}$
$\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}$
The above matrix is equal to the identity matrix.
Hence, it is proved that AA^-1 = A^-1A = I