# 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 = I_{n} where I_{n }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 = 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= ** **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=I

_{n}or notAB =

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(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 = , 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^{-1}A^{-1}**

**Proof:**

|A| ≠0, |B| ≠0

So, |AB| ≠0

Let a matrix C = B

^{-1}A^{-1}(AB)C = (AB)B

^{-1}A^{-1}(AB)C = A(BB

^{-1})A^{-1}(AB)C = AI

_{n}A^{-1}(AB)C = AA

^{-1}(AB)C = I

_{n}C(AB) = B

^{-1}A^{-1}(AB)C(AB) = B

^{-1}A^{-1}ABC(AB) = B

^{-1}BC(AB) = I

_{n}Since (AB)C = C(AB) = I

_{n}Hence, C is inverse of (AB)

So (AB)

^{-1}= B^{-1}A^{-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

^{-1}A^{-1}=B

^{-1}A^{-1}=B

^{-1}A^{-1}= …….. (2)From (1) and (2), you can see that (AB)

^{-1}= B^{-1}A^{-1}Hence, it is proved that (AB)

^{-1}= B^{-1}A^{-1}

### Properties of inverse of a square matrix

**1. (A ^{-1})^{-1 }= A**

**Proof:**

If A is an invertible matrix then

AA

^{-1}= ITaking inverse on both sides

(AA

^{-1})^{-1}= I^{-1}(A

^{-1})^{-1}A^{-1}= I [from theorem 2 (AB)^{-1}= B^{-1}A^{-1}]Multiplying by A on both sides

(A

^{-1})^{-1}A^{-1}A = IA(A

^{-1})^{-1}I = A(A

^{-1})^{-1}= AHence, it is proved that (A

^{-1})^{-1}= A

**Example:**

Let A =

|A| = 12 – 2 = 10

adj A =

A

^{-1}= adj A /|A|A

^{-1}=(A

^{-1})^{-1}= adj (A^{-1}) / |A^{-1}|adj(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.** **(A _{1}A_{2}A_{3}………..A_{n})^{-1} = A_{n}^{-1}A_{n-1}^{-1}……….A_{2}^{-1}A_{1}^{-1}**

**You can also write it:**

**(AB) ^{-1} = A^{-1}B^{-1}**

**(ABC) ^{-1} = A^{-1}B^{-1}C^{-1}**

**Proof: **

This can be proved by mathematical induction

for n = 2

(A

_{1}A_{2})^{-1}= A_{2}^{-1}A_{1}^{-1}……….(1)This statement is true. [by theorem 2]

Let this is true for n = k

(A

_{1}A_{2}A_{3}……….A_{k})^{-1}= A_{k}^{-1}…………A_{2}^{-1}A_{1}^{-1}……..(2)For n = k+1, you have to prove this.

(A

_{1}A_{2}A_{3}……….A_{k}A_{k+1})^{-1}=((A

_{1}A_{2}A_{3}………A_{k})A_{k+1})^{-1}=((A

_{k}^{-1}…………A_{2}^{-1}A_{1}^{-1})A_{k+1})^{-1}=(A

_{k+1})^{-1}(A_{k}^{-1}…………A_{2}^{-1}A_{1}^{-1}) [using theorem 2]= A

_{k+1}^{-1}A_{k}^{-1}………….A_{2}^{-1}A_{1}^{-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

^{-1}A^{-1}=B

^{-1}A^{-1}=B

^{-1}A^{-1}= …….. (2)From (1) and (2), you can see that (AB)

^{-1}= B^{-1}A^{-1}

Hence, it is proved that (AB)^{-1}= B^{-1}A^{-1}

**3. AA ^{-1}= A^{-1}A = I_{n}**

**Proof: **

A matrix is invertible if AA

^{-1}= IMultiply by A on both sides

AAA

^{-1}= AIAI = A

Multiplying by A

^{-1}on both sidesA

^{-1}AI = A^{-1}AI = A

^{-1}AHence, it is proved that AA

^{-1}= I = A^{-1}A

**Example: **

Let A =

|A| = 3 + 2 = 5

adj A =

A

^{-1}= adj A \|A|A

^{-1}=Now to prove AA

^{-1}= A^{-1}A = I

Taking left hand sideAA

^{-1}The above matrix is equal to the identity matrix.

Now taking right sideA

^{-1}AThe above matrix is equal to the identity matrix.

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

**More Properties:**

- (A
^{T})^{-1}=(A^{-1})^{T} - (kA)
^{-1}= (1/k)A^{-1} - AB = I
_{n}, where A and B are inverse of each other. - If A is square matrix where n > 0, then (A
^{-1})^{n }= A^{-n}

## Please

Loginto comment...