Invertible Matrix

An invertible matrix is a square matrix whose inverse exists.Let A be a square matrix of order n × n. The matrix A is said to be invertible if there exists another matrix B of the same order such that,

AB = BA = Iₙ
where Iₙ is the identity matrix of order n × n.
In this case, B is called the inverse of A and is denoted by A⁻¹.

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

Solution:

Given matrices,
A = \begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix}
B = \begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix}
Multiplying A and B
⇒ 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 .......... (i)
Similarly, Multiplying B and 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............ (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 = \bold{\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 = \bold{\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
⇒ |A| ≠ 0
Hence. matrix A is invertible.
Now, check whether B is invertible matrix of A we have to prove,
AB = BA = I_n
⇒ 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...(i)
And, 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...(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

Inverse can only be calculated for Square Matrix whose determinant is non-zero.
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 only if B = A⁻¹.
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

The Proof forDet (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

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

Adjoint of a Matrix
Determinant of a Matrix

Solved Examples on Invertible Matrix

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

Given,
A = \begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix} ...(i)
⇒ |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} ...(ii)
From (i) and (ii) (A^-1)^-1 = A
Hence, Proved

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

Given Matrix,
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
LHS
⇒ LHS= AA^-1
⇒ LHS= \begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}\frac{1}{5}\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix}
⇒ LHS= \frac{1}{5}\begin{bmatrix} 3 +2 &-2 +2\\ -3+3 &2+3 \end{bmatrix}
⇒ LHS= \frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix}
⇒ LHS= \begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}
⇒ LHS= I ...(i)
RHS
⇒ RHS= A^-1A
⇒ RHS= \frac{1}{5} \begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix} \begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}
⇒ RHS= \frac{1}{5}\begin{bmatrix} 3+2 & 6-6\\ 1-1&2+3 \end{bmatrix}
⇒ RHS= \frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix}
⇒ RHS= \begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}
⇒ RHS= I...(ii)
From (i) and (ii) AA^-1 = A^-1A = I
Hence, Proved

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

Given Matrix,
A = \begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}
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} .......... (i)
We know that,
A^-1 = Adjoint 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} ........ (ii)
From (ii) and (ii), you can see that (AB)^-1 = B^-1A^-1
Hence, Proved

Matrix Inversion Methods

Invertible Matrix Theorem

Theorem 1

Theorem 2

Invertible Matrix Properties

Invertible Matrix Determinant

Proof

Applications of Invertible Matrix

Proof For Properties of Invertible Matrix

Related Articles:

Solved Examples on Invertible Matrix

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