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Inverse of a Matrix

Last Updated : 07 Mar, 2025
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The inverse of a Matrix "A", denoted as A−1, is another matrix that, when multiplied by A, results in the identity matrix 𝜤. In mathematical terms:

A × A-1 = A-1 × A = 𝜤

Where 𝜤 is the identity matrix, a square matrix in which all the elements of the principal diagonal are 1, and all other elements are 0.

Note: Not all matrices have an inverse. A matrix must be square (same number of rows and columns) and must be non-singular (its determinant is not zero) to have an inverse.

The inverse of a matrix is obtained by dividing the adjugate(also called adjoint) of the given matrix by the determinant of the given matrix.

inverse of matrix

The inverse of a matrix is another matrix that, when multiplied by the given matrix, yields the multiplicative identity.

The terminology listed below can help you grasp the inverse of a matrix more clearly and easily.

Terms DefinitionFormula/ProcessExample with Matrix A
MinorThe minor of an element in a matrix is the determinant of the matrix formed by removing the row and column of that element.For element aij , remove the ith row and jth column to form a new matrix and find its determinant.Minor of a11 is the determinant of

A = \begin{bmatrix}5 & 6\\ 6 & 7\end{bmatrix}

CofactorThe cofactor of an element is the minor of that element multiplied by (-1)i+j , where i and j are the row and column indices of the element.Cofactor of aij = (-1)i+j Minor of aij Cofactor of a11 = (-1)1+1 × Minor of a11 = Minor of a11
DeterminantThe determinant of a matrix is calculated as the sum of the products of the elements of any row or column and their respective cofactors.For a row (or column), sum up the product of each element and its cofactor.Determinant of A = a11 ​× Cofactor of a11 +a12 × Cofactor of a12​ +a13 × Cofactor of a13​.
AdjointThe adjoint of a matrix is the transpose of its cofactor matrix.Create a matrix of cofactors for each element of the original matrix and then transpose it.Adjoint of A is the transpose of the matrix formed by the cofactors of all elements in A.

Singular Matrix

A matrix whose value of the determinant is zero is called a singular matrix, i.e. any matrix A is called a singular matrix if |A| = 0. The inverse of a singular matrix does not exist.

Non-Singular Matrix

A matrix whose value of the determinant is non-zero is called a non-singular matrix, i.e. any matrix A is called a non-singular matrix if |A| ≠ 0. The inverse of a non-singular matrix exists.

Identity Matrix

A square matrix in which all the elements are zero except for the principal diagonal elements is called the identity matrix. It is represented using 𝜤. It is the identity element of the matrix as for any matrix A,

A × 𝜤 = A

An example of an Identity matrix is,

𝜤3×3 = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{bmatrix}

This is an identity matrix of order 3×3.

How to Find Inverse of a Matrix?

There are Two-ways to find the Inverse of a matrix in mathematics:

  • Using Matrix Formula
  • Using Inverse Matrix Methods

Using the Matrix Formula

The inverse of matrix A, that is A-1 is calculated using the inverse of matrix formula, which involves dividing the adjoint of a matrix by its determinant.

A^{-1}=\frac{\text{Adj A}}{|A|}  

where,

  • adj A = adjoint of the matrix A, and 
  • |A| = determinant of the matrix A.
inverse-of-a-matrix

Note: This formula only works on Square matrices.

To find the inverse of the matrix using inverse of a matrix formula, follow these steps.

Step 1: Determine the minors of all A elements.

Step 2: Next, compute the cofactors of all elements and build the cofactor matrix by substituting the elements of A with their respective cofactors.

Step 3: Take the transpose of A's cofactor matrix to find its adjoint (written as adj A).

Step 4: Multiply adj A by the reciprocal of the determinant of A.

Now, for any non-singular square matrix A,

A-1 = 1 / |A| × Adj (A)

Example: Find the inverse of the matrix A=\left[\begin{array}{ccc}4 & 3 & 8\\6 & 2 & 5\\1 & 5 & 9\end{array}\right]using the formula.

We have, A=\left[\begin{array}{ccc}4 & 3 & 8\\6 & 2 & 5\\1 & 5 & 9\end{array}\right]

Find the value of determinant of the matrix.

|A| = 4(18–25) – 3(54–5) + 8(30–2)

⇒ |A| = 49

Find the adjoint of matrix A by computing the cofactors of each element and then getting the cofactor matrix's transpose.

adj A = \left[\begin{array}{ccc}-7 & 13 & 1\\-49 & 28 & 28\\28 & -17 & -10\end{array}\right]

So, the inverse of the matrix is,

A–1\frac{1}{49}\left[\begin{array}{ccc}-7 & 13 & 1\\-49 & 28 & 28\\28 & -17 & -10\end{array}\right]

⇒ A–1\left[\begin{array}{ccc}- \frac{1}{7} & \frac{13}{49} & - \frac{1}{49}\\-1 & \frac{4}{7} & \frac{4}{7}\\\frac{4}{7} & - \frac{17}{49} & - \frac{10}{49}\end{array}\right]

Inverse Matrix Methods to Find Inverse Of a Matrix

There are two Inverse matrix methods to find matrix inverse:

  1. Determinant Method
  2. Elementary Transformation Method

Method 1: Determinant Method

The most important method for finding the matrix inverse is using a determinant.

The inverse matrix is also found using the following equation:

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

where,

  • adj(A) is the adjoint of a matrix A, and
  • det(A) is the determinant of a matrix A.

For finding the adjoint of a matrix A the cofactor matrix of A is required. Then adjoint (A) is the transpose of the Cofactor matrix of A i.e.,

adj (A) = [Cij]T

  • For the cofactor of a matrix i.e., Cij, we can use the following formula:

Cij = (-1)i+j det (Mij)

Where Mij refers to the (i, j)th minor matrix when ith row and jth column are removed.

Method 2: Elementary Transformation Method

Follow the steps below to find an Inverse matrix by the elementary transformation method.

Step 1: Write the given matrix as A = IA, where I is the identity matrix of the order same as A.

Step 2: Use the sequence of either row operations or column operations till the identity matrix is achieved on the LHS also use similar elementary operations on the RHS such that we get I = BA. Thus, the matrix B on RHS is the inverse of matrix A.

Step 3: Make sure we either use Row Operation or Column Operation while performing elementary operations.

We can easily find the inverse of the 2 × 2 Matrix using the elementary operation. Let's understand this with the help of an example.

Example: Find the inverse of the 2 × 2, A =  \begin{bmatrix}2 & 1\\ 1 & 2\end{bmatrix} using the elementary operation.

Solution:

Given: A = 𝜤A

\begin{bmatrix}2 & 1\\ 1 & 2\end{bmatrix}~=~\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}~×~\begin{bmatrix}2 & 1\\ 1 & 2\end{bmatrix}

Now, R1 ⇢ R1/2
\begin{bmatrix}1 & 1/2\\ 1 & 2\end{bmatrix}~=~\begin{bmatrix}1/2 & 0\\ 0 & 1\end{bmatrix}~×~A

R2 ⇢ R2 - R1
\begin{bmatrix}1 & 1/2\\ 0 & 3/2\end{bmatrix}~=~\begin{bmatrix}1/2 & 0\\ -1/2 & 1\end{bmatrix}~×~A

R2 ⇢ R2 × 2/3
\begin{bmatrix}1 & 1/2\\ 0 & 1\end{bmatrix}~=~\begin{bmatrix}1/2 & 0\\-1/3 & 2/3\end{bmatrix}~×~A

R1 ⇢ R1 - R2/2
\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}~=~\begin{bmatrix}2/3 & -1/3\\ -1/3 & 2/3\end{bmatrix}~×~A

Thus, the inverse of the matrix A =  \begin{bmatrix}2 & 1\\ 1 & 2\end{bmatrix} is A-1\begin{bmatrix}2/3 & -1/3\\ -1/3 & 2/3\end{bmatrix}

2 × 2 Matrix Inverse

The inverse of the 2×2 matrix can also be calculated using the shortcut method apart from the method discussed above. Let's consider an example to understand the shortcut method to calculate the inverse of 2 × 2 Matrix.

For given matrix A = \begin{bmatrix}a & b\\ c & d\end{bmatrix}

We know, |A| = (ad - bc)

and adj A = \begin{bmatrix}d & -b\\ -c & a\end{bmatrix}

then using the formula for inverse

A-1 =  (1 / |A|) × Adj A

⇒ A-1[1 / (ad - bc)] × \begin{bmatrix}d & -b\\ -c & a\end{bmatrix}

Thus, the inverse of the 2 × 2 matrix is calculated.

3 × 3 Matrix Inverse

Let us take any 3×3 Matrix:

A = \begin{bmatrix}a & b & c\\ l & m & n\\ p & q & r\end{bmatrix}

The inverse of the 3×3 matrix is calculated using the inverse matrix formula

A-1 = (1 / |A|) × Adj A

The inverse of Diagonal Matrices

If you have a diagonal matrix, finding the inverse is quite easy. For a matrix:

\begin{bmatrix}a & 0 & 0\\ 0 & b & 0\\ 0 & 0 & c\end{bmatrix}The inverse

The inverse is:

\begin{bmatrix}\frac{1}{a} & 0 & 0\\ 0 & \frac{1}{b} & 0\\ 0 & 0 & \frac{1}{c}\end{bmatrix}

Simply take the reciprocal of each diagonal element.

Note: Inverse of Orthogonal Matrices: 
If A is an orthogonal matrix (i.e., AT = A−1), its inverse is simply its transpose.

Determinant of Inverse Matrix

The determinant of the inverse matrix is the reciprocal of the determinant of the original matrix. i.e., 

det(A-1) = 1 / det(A)

The proof of the above statement is discussed below:

det(A × B) = det (A) × det(B)  (already know)
⇒ A × A-1 = I  (by Inverse matrix property)
⇒ det(A × A-1) = det(I)
⇒ det(A) × det(A-1) = det(I)     [ but, det(I) = 1]
⇒ det(A) × det(A-1) = 1
⇒ det(A-1) = 1 / det(A)

Hence, Proved.

Properties of Inverse Matrix

The inverse matrix has the following properties:

  • For any non-singular matrix A, (A-1)-1 = A
  • For any two non-singular matrices A and B, (AB)-1 = B-1A-1
  • The inverse of a non-singular matrix exists, for a singular matrix, the inverse does not exist.
  • For any nonsingular A, (AT)-1 = (A-1)T

Related Articles:

Matrix Inverse Solved Examples

Let's solve some example questions on the Inverse of the Matrix.

Example 1: Find the inverse of the matrix \bold{A=\left[\begin{array}{ccc}2 & 3 & 1\\1 & 1 & 2\\2 & 3 & 4\end{array}\right]} using the formula.

Solution:

We have,

A=\left[\begin{array}{ccc}2 & 3 & 1\\1 & 1 & 2\\2 & 3 & 4\end{array}\right]

Find the adjoint of matrix A by computing the cofactors of each element and then getting the cofactor matrix's transpose.

adj A = \left[\begin{array}{ccc}-2 & -9 & 5\\0 & 6 & -3\\1 & 0 & -1\end{array}\right]

Find the value of determinant of the matrix.

|A| = 2(4–6) – 3(4–4) + 1(3–2)

= –3

So, the inverse of the matrix is,

A–1\frac{1}{-3}\left[\begin{array}{ccc}-2 & -9 & 5\\0 & 6 & -3\\1 & 0 & -1\end{array}\right]

      = \left[\begin{array}{ccc}\frac{2}{3} & 3 & - \frac{5}{3}\\0 & -2 & 1\\- \frac{1}{3} & 0 & \frac{1}{3}\end{array}\right]

Example 2: Find the inverse of the matrix A =\left[\begin{array}{ccc}6 & 2 & 3\\0 & 0 & 4\\2 & 0 & 0\end{array}\right]      

Solution:

We have,

A=\left[\begin{array}{ccc}6 & 2 & 3\\0 & 0 & 4\\2 & 0 & 0\end{array}\right]

Find the adjoint of matrix A by computing the cofactors of each element and then getting the cofactor matrix's transpose.

adj A = \left[\begin{array}{ccc}0 & 0 & 8\\8 & -6 & -24\\0 & 4 & 0\end{array}\right]

Find the value of determinant of the matrix.

|A| = 6(0–4) – 2(0–8) + 3(0–0)

     = 16

So, the inverse of the matrix is,

A–1\frac{1}{16}\left[\begin{array}{ccc}0 & 0 & 8\\8 & -6 & -24\\0 & 4 & 0\end{array}\right]

      = \left[\begin{array}{ccc}0 & 0 & \frac{1}{2}\\\frac{1}{2} & - \frac{3}{8} & - \frac{3}{2}\\0 & \frac{1}{4} & 0\end{array}\right]

Example 3: Find the inverse of the matrix A=\bold{\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & 4\\0 & 0 & 1\end{array}\right]      }      using the formula.

Solution:

We have,

A=\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & 4\\0 & 0 & 1\end{array}\right]

Find the adjoint of matrix A by computing the cofactors of each element and then getting the cofactor matrix's transpose.

adj A = \left[\begin{array}{ccc}1 & -2 & 5\\0 & 1 & -4\\0 & 0 & 1\end{array}\right]

Find the value of determinant of the matrix.

|A| = 1(1–0) – 2(0–0) + 3(0–0)

= 1

So, the inverse of the matrix is,

A–1\frac{1}{1}\left[\begin{array}{ccc}1 & -2 & 5\\0 & 1 & -4\\0 & 0 & 1\end{array}\right]

\left[\begin{array}{ccc}1 & -2 & 5\\0 & 1 & -4\\0 & 0 & 1\end{array}\right]

Example 4: Find the inverse of the matrix A=\bold{\left[\begin{array}{ccc}1 & 2 & 3\\2 & 1 & 4\\3 & 4 & 1\end{array}\right]      }      using the formula.

Solution:

We have,

A=\left[\begin{array}{ccc}1 & 2 & 3\\2 & 1 & 4\\3 & 4 & 1\end{array}\right]

Find the adjoint of matrix A by computing the cofactors of each element and then getting the cofactor matrix's transpose.

adj A = \left[\begin{array}{ccc}-15 & 10 & 5\\10 & -8 & 2\\5 & 2 & -3\end{array}\right]

Find the value of determinant of the matrix.

|A| = 1(1–16) – 2(2–12) + 3(8–3)

= 20

So, the inverse of the matrix is,

A–1\frac{1}{20}\left[\begin{array}{ccc}-15 & 10 & 5\\10 & -8 & 2\\5 & 2 & -3\end{array}\right]

      = \left[\begin{array}{ccc}- \frac{3}{4} & \frac{1}{2} & \frac{1}{4}\\\frac{1}{2} & - \frac{2}{5} & \frac{1}{10}\\\frac{1}{4} & \frac{1}{10} & - \frac{3}{20}\end{array}\right]


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