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Invertible Matrices

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 = In where In is 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 =

and B =

On multiplying A and B you get,

AB =

AB =

AB =

AB = I ………. (1)

Similarly, you can get BA by multiplying matrix B and matrix A.

BA =

BA =

BA =

BA = I………… (2)

From (1) and (2), you can see that AB = BA = In

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:

1. Least-squares or Regression
2. Simulations
3. 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’:

1. Gaussian Elimination
2. Newton’s Method
3. Cayley-Hamilton Method
4. Eigen Decomposition Method

Example: Check whether matrix A=  is invertible or not. And if A is invertible then check whether matrix B =  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=In or not

AB =

AB =

AB =

AB =

AB =

AB = I

BA =

BA =

BA =

BA =

BA =

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(In) = C ……..(3)

From (2) you have

(CA)B = In(B) = B ……. (4)

Since, C(AB) = (CA)B [Associativity Law]

So, C = B

Hence, it is proved.

Example:

Let A = , B = , and C =

If both matrices B and C are inverse of matrix A then,

AB = BA = I and

AC = CA = I

Taking AB = I

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 = ………… (5)

Now, consider AC = I

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 = ………..(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 = AInA-1

(AB)C = AA-1

(AB)C = In

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

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

C(AB) = B-1B

C(AB) = In

Since (AB)C = C(AB) = In

Hence, C is inverse of (AB)

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

Example:

Let A = and B =

AB =

AB =

AB =

(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 =

B-1

A =

A-1

B-1A-1

B-1A-1

B-1A-1 …….. (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 =

|A| = 12 – 2 = 10

A-1

(A-1)-1 = adj (A-1) / |A-1|

|A-1| = (12 – 10)/100

|A-1| = 1/10

(A-1)-1

(A-1)-1

From above you can see that (A-1)-1 = A

2. (A1A2A3………..An)-1 = An-1An-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……….Ak)-1 = Ak-1…………A2-1A1-1……..(2)

For n = k+1, you have to prove this.

(A1A2A3……….AkAk+1)-1

=((A1A2A3………Ak)Ak+1)-1

=((Ak-1…………A2-1A1-1)Ak+1)-1

=(Ak+1)-1 (Ak-1…………A2-1A1-1)    [using theorem 2]

= Ak+1-1Ak-1………….A2-1A1-1

Hence, it is proved.

Example:

Suppose there are two matrices A and B,

Let A =

and B =

AB =

AB =

AB =

(AB)-1 ………. (1)

Inverse of a matrix can be obtained by the given formula

A-1 = adjoint of matrix A/ |A|

B =

B-1

A =

A-1

B-1A-1

B-1A-1

B-1A-1…….. (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 = In

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 =

|A| = 3 + 2 = 5

A-1

Now to prove AA-1 = A-1A = I

Taking left hand side

AA-1

The above matrix is equal to the identity matrix.

Now taking right side

A-1A

The above matrix is equal to the identity matrix.

Hence, it is proved that AA-1 = A-1A = I

More Properties:

• (AT)-1 =(A-1)T
• (kA)-1 = (1/k)A-1
• AB = In, where A and B are inverse of each other.
• If A is square matrix where n > 0, then (A-1)n = A-n

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