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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 = 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 = \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 = 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= \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=In 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(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 = \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 = 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 =\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)-110*\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………..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 = \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 = 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 = \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

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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