Invertible Matrix

Invertible matrices are defined as the matrix whose inverse exists. We define a matrix as the arrangement of data in rows and columns, if any matrix has m rows and n columns then the order of the matrix is m × n where m and n represent the number of rows and columns respectively. 

We define invertible matrices as square matrices whose inverse exists. They are non-singular matrices as their determinant exists. There are various methods to calculate the inverse of the matrix.

In this article, we will learn about, What are Invertible Matrices? Invertible Matrices Examples, Invertible Matrix Theorems, Invertible Matrix Determinant, and others in detail.

What is Invertible Matrix?

Invertible matrices are defined as the matrix whose inverse exists. We can also say that invertible matrices are the matrix for which inversion operations exist. An invertible matrix is a square matrix as the inverse of only a square matrix exists. The order of the invertible matrix is of the form, n × n. Let A be any square matrix of order n × n if there exists a matrix of order B of order n × n, such that,

AB = BA = I_n

where 
I_n is the identity matrix of order n × n

The basic condition of the invertible matrices is discussed in the image below.

Invertible Matrix Example

Example: Check if A = [Tex]\bold{\begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix}}     [/Tex] is an invertible matrix and B = [Tex]\bold{\begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix}}     [/Tex] is its inverse.

Solution:

Given matrices,

A = [Tex]\begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix} [/Tex]

B = [Tex]\begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix} [/Tex]

Multiplying A and B

⇒ AB = [Tex]\begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix} \begin{bmatrix} 5&-6 \\ -4&5 \end{bmatrix} [/Tex]

⇒ AB = [Tex]\begin{bmatrix} 25-24 &-30+30 \\ 20-20&-24+25 \end{bmatrix} [/Tex]

⇒ AB = [Tex]\begin{bmatrix}1 &0 \\ 0&1 \end{bmatrix} [/Tex]

⇒ AB = I ………. (i)

Similarly, Multiplying B and A

⇒ BA =[Tex] \begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix} \begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix} [/Tex]

⇒ BA = [Tex]\begin{bmatrix} 25-24 &30-30 \\ -20+20&-24+25 \end{bmatrix} [/Tex]

⇒ BA =[Tex] \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix} [/Tex]

⇒ BA = I………… (ii)

From (i) and (ii), we see that AB = BA = I_n

Hence, A is an invertible matrix and the inverse of matrix A is matrix B. 

This can be written as A^-1 = B.

If B is the inverse matrix for A then also, A is the 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, for a matrix A to be invertible if and only if |A| ≠ 0.

Matrix Inversion Methods

There are various methods to find the inversion of the matrix. Using the following methods we can find the other matrix the ‘B’ matrix which is the inverse of the matrix ‘A’:

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

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

⇒ |A| ≠ 0

Hence. matrix A is invertible.

Now, check whether Bis invertible matrix of A we have to prove,

AB = BA = I_n

⇒ AB = [Tex] \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} [/Tex]

⇒ AB = [Tex]\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} [/Tex]

⇒ AB = [Tex]\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} [/Tex]

⇒ AB = [Tex]\frac{-1}{2} \begin{bmatrix} -2&0  &0\\ 0&-2  &0 \\ 0& 0& -2 \end{bmatrix} [/Tex]

⇒ AB = [Tex]\begin{bmatrix} 1&0  &0\\ 0&1  &0 \\ 0& 0& 1 \end{bmatrix} [/Tex]

⇒ AB = I…(i)

And, BA =[Tex] \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} [/Tex]

⇒ BA = [Tex]\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} [/Tex]

⇒ BA = [Tex]\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} [/Tex]

⇒ BA = [Tex]\frac{-1}{2} \begin{bmatrix} -2&0  &0\\ 0&-2  &0 \\ 0& 0& -2 \end{bmatrix} [/Tex]

⇒ BA = [Tex]\begin{bmatrix} 1&0  &0\\ 0&1  &0 \\ 0& 0& 1 \end{bmatrix} [/Tex]

⇒ BA = I…(ii)

From (i) and (ii) 

AB = BA = I

Hence, matrix B is inverse of matrix A

Invertible Matrix Theorem

There are two very basic invertible matrix theorems that are used in solving various problems of matrices, which are listed as follows:

Theorem 1

Statement: Every invertible matrix has a unique inverse.

Proof: 

Let ‘A’ be an n×n invertible matrix.

Let us consider B and C to be two inverses of A.

Then,

AB = BA = I . . .(i)

and AC = CA = I . . . (ii)

From (i) you have

C(AB) = C(I_n) = C ……..(iii)

From (ii) you have

(CA)B = I_n(B) = B ……. (iv)

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

So, C = B

Hence, proved.

Theorem 2

Statement: For two matrices A and B of the same order and if their multiplication AB exists. Then (AB)^-1= B^-1A^-1

Proof:

Now, |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 the inverse of (AB)

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

Hence, Proved

Some other theorems of the Invertible matrices are if A is the invertible matrix of order n×n then the equivalence condition of the matrix a to be invertible is,

• A is row-equivalent to the Identity matrix I_n of order n×n.
• The only trivial solution of the equation Ax = 0 is, x = 0
• Linear transformation x: → Ax is a one-one transformation.
• There exists a matrix C of order n×n such that CA = I_n exists.
• There exists a matrix D of order n×n such that AD = I_n exists.
• The transpose of the matrix A, A^T is invertible.
• The rank of matrix A is n.
• The determinant of matrix A is always non-zero.

Invertible Matrix Properties

Invertible matrices have various properties and some of the important properties of invertible matrices are listed below,

• Inverse can only be calculated for Square Matrix whose determinant is non.
• For the square matrix also the inverse of only the non-singular matrix exists. Non-singular matrices are the matrices where the determinant is non-zero.
• For any non-singular matrix A, (A^T)^-1 = (A^-1)T where A^T represents the Transpose matrix of A.
• For any two invertible matrices A and B, AB = I_n where I_n is the identity matrix.
• If the inverse of any matrix A exists then x = A^-1B is the solution of the equation, Ax = B
• Det (A^-1) = (Det A)^-1
• (cA)^-1 = 1/c.A^-1

Invertible Matrix Determinant

We define that for any square matrix A the determinant of the inverse of the square matrix A is the reciprocal of the determinant of the square matrix, i.e.

Det (A^-1) = 1/Det(A)

Proof of Invertible Matrix Determinant

The Proof for Det (A^-1) = 1/Det(A) is discussed below,

We know that,

Det(A × B) = Det (A) × Det(B)

A × A^-1 = I_n          (property of invertible matrix)

⇒ Det(A ×A^-1) = Det(I_n)

⇒ Det(A) × Det(A^-1) = Det(I_n)           {Det(I_n) = 1}

⇒ Det(A) × Det(A^-1) = 1

⇒ Det(A^-1) = 1 / Det(A)

Hence, proved.

Applications of Invertible Matrix

Invertible matrix has various applications and some of the important applications of invertible matrices are,

• They are  used in cryptography to design and solve complex codes
• They are used in the 3-D automation of various objects.
• They are used to encrypt and decrypt messages.
• They are used in programming well-optimized solutions to various problems.
• They are used in the image processing of cameras.

Proof For Properties of Invertible Matrix

There are various properties of invertible matrices, some of which are discussed as follows:

•  (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

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

• 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

Read More,

• Adjoint of a Matrix
• Determinant of a Matrix

Solved Examples on Invertible Matrix

Example 1: If [Tex]A = \begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}         [/Tex] prove, (A^-1)^-1 = A

Given,

A = [Tex]\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}          [/Tex]…(i)

⇒ |A| = 12 – 2 = 10

⇒ adj A = [Tex]\begin{bmatrix} 4 &-1 \\ -2 & 3 \end{bmatrix} [/Tex]

⇒ A^-1 = adj A /|A|

⇒ A^-1 = [Tex]\frac{1}{10}\begin{bmatrix} 4 &-1 \\ -2 & 3 \end{bmatrix} [/Tex]

⇒ (A^-1)^-1 = adj (A^-1) / |A^-1|

⇒ adj(A^-1) = [Tex]\frac{1}{10}\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix} [/Tex]

⇒ |A^-1| = (12 – 10)/100

⇒ |A^-1| = 1/10

⇒ (A^-1)^-1 = [Tex]10×\frac{1}{10}\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix} [/Tex]

⇒ (A^-1)^-1 = [Tex]\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}          [/Tex]…(ii)

From (i) and (ii) (A^-1)^-1 = A

Hence, Proved

Example 2: If A = [Tex]\begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}          [/Tex] Prove, AA^-1 = A^-1A = I

Given Matrix,

A = [Tex]\begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix} [/Tex]

⇒ |A| = 3 + 2 = 5

⇒ adj A = [Tex]\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix} [/Tex]

⇒ A^-1= adj A \|A|

⇒ A^-1 = [Tex]\frac{1}{5}\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix} [/Tex]

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

LHS

⇒ LHS= AA^-1

⇒ LHS= [Tex]\begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}\frac{1}{5}\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix} [/Tex]

⇒ LHS= [Tex]\frac{1}{5}\begin{bmatrix} 3 +2 &-2 +2\\ -3+3 &2+3 \end{bmatrix} [/Tex]

⇒ LHS= [Tex]\frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix} [/Tex]

⇒ LHS= [Tex]\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix} [/Tex]

⇒ LHS= I …(i)

RHS

⇒ RHS= A^-1A

⇒ RHS= [Tex]\frac{1}{5} \begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix} \begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix} [/Tex]

⇒ RHS= [Tex]\frac{1}{5}\begin{bmatrix} 3+2 & 6-6\\ 1-1&2+3 \end{bmatrix} [/Tex]

⇒ RHS= [Tex]\frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix} [/Tex]

⇒ RHS= [Tex]\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix} [/Tex]

⇒ RHS= I…(ii)

From (i) and (ii) AA^-1 = A^-1A = I

Hence, Proved

Example 3: If A  = [Tex]\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}  and B = \begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}         [/Tex] Prove, (AB)^-1 = B^-1A^-1

Given Matrix,

A = [Tex]\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}   [/Tex]

B =[Tex]\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix} [/Tex]

⇒ AB = [Tex]\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix} \begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix} [/Tex]

⇒ AB = [Tex]\begin{bmatrix} 3+4 &0+8 \\ 6+0& 0 \end{bmatrix} [/Tex]

⇒ AB = [Tex]\begin{bmatrix} 7 &8 \\ 6& 0 \end{bmatrix} [/Tex]

⇒ (AB)^-1 = [Tex]\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}              [/Tex] ………. (i)

We know that,

A^-1 = Adjoint A/ |A|

⇒ B = [Tex]\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix} [/Tex]

⇒ B^-1 = [Tex]\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix} [/Tex]

A = [Tex]\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix} [/Tex]

⇒ A^-1 = [Tex]\frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix} [/Tex]

⇒ B^-1A^-1 = [Tex]\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix} \frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix} [/Tex]

⇒ B^-1A^-1 = [Tex]\frac{-1}{48}\begin{bmatrix} 0 &-8+0 \\ 0-6& 4+3 \end{bmatrix} [/Tex]

⇒ B^-1A^-1 = [Tex]\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}               [/Tex]…….. (ii)

From (ii) and (ii), you can see that (AB)^-1 = B^-1A^-1

Hence, Proved

Invertible Matrix – FAQs

What are Invertible Matrices

Invertible matrices are defined as matrices whose inverse exists. They are a non-singular, square matrix of order n×n. If A is an invertible matrix and its inverse is  B, then 

A×B = I_n

where I_n is the inverse matrix.

What are Properties of an Invertible Matrix?

Some of the important properties of invertible matrices are,

• The inverse of any matrix is unique.
• x = A^-1B is the solution of the equation, Ax = B
• Det (A^-1) = (Det A)^-1
• (cA)^-1 = 1/c.A^-1

What is the condition for an Invertible Matrix?

For any square matrix A, the condition for the matrix to be invertible is, there exists a unique square matrix B such that,

A×B = I_n

What is a Non-Invertible Matrix?

A matrix whose inverse does not exists are called a non-invertible matrix. All the singular matrices and the matrix which are not square are non-invertible matrices.

What is Inverse Matrix Theorem?

The inverse matrix theorems are discussed below,

• Every invertible matrix has a unique inverse.
• For two matrices A and B of the same order and if their multiplication AB exists. Then (AB)^-1= B^-1A^-1.

Can a Matrix be Invertible if its determinant is 0?

No, a matrix can not be invertible if its determinant is 0. As we know that a matrix is invertible if the inverse of the matrix exists and the inverse of matrix A is calculated as,

A^-1 = adj A/|A|

And if the determinant of the matrix is zero then its inverse does not exist and hence it is invertible.

What is Invertibility of a Matrix?

A square matrix is invertible iff its determinant is non-zero. Let us suppose we take a square matrix A of order n×n, then the inverse of the matrix exist only when determinant of matrix A is non-zero, i.e. |A| ≠ 0.

How do you Determine the Invertibility of a Matrix?

For any matrix if its determinant is zero then the matrix is not invertible or else the matrix is invertible.

What do Eigenvalues say about Invertibility?

For an invertible matrix we cannot have an eigenvalue equal to zero.

Is A 2×2 matrix always invertible?

No, a 2×2 matrix is not always invertible. It is only invertible if the determinant of the 2×2 matrix is non-zero. If the determinant of the 2×2 matrix is zero, then the matrix is singular and does not have an inverse.

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