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# What is the Inverse of a Matrix – Definition, Formula & Examples

February 18, 2023

You studied the multiplicative inverse of a number. A multiplicative inverse of a number $a$ is a number $b$, which when multiplied with the number results in a multiplicative identity(i.e., $1$). Mathematically, it can be stated as $b$ is called a multiplicative inverse of $a$, if $a \times b = b \times a = 1$.

Similarly, a matrix $\text{B}$ is called a multiplicative inverse of a matrix $\text{A}$, if $\text{A} \times \text{B} = \text{I}$, where $\text{I}$ is the identity matrix.

Let’s understand what is the inverse of a matrix and what is its formula with examples.

## What is the Inverse of a Matrix?

The inverse of a matrix is another matrix, which on multiplication with the given matrix gives the multiplicative identity. For a matrix, $\text{A}$ its inverse is denoted by $\text{A}^{-1}$, and $\text{A} \times \text{A}^{-1} = \text{A}^{-1} \times \text{A} = \text{I}$, where $\text{I}$ is the identity matrix.

The matrix whose determinant is non-zero and for which the inverse matrix can be calculated is called an invertible matrix. For example, the inverse of $\text{A} = \begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix}$ is $\text{A}^{-1} = \frac{1}{3} \begin{bmatrix} 0 & -1 \\ 3 & 2 \end{bmatrix}$.

The product of  $\text{A}$ and $\text{A}^{-1}$ is an identity matrix $\text{I}$.

• Premultiplying $\text{A}$ with $\text{A}^{-1}$:

$\text{A} \times \text{A}^{-1} = \begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix} \times \frac{1}{3} \begin{bmatrix} 0 & -1 \\ 3 & 2 \end{bmatrix}$

$= \frac{1}{3} \begin{bmatrix} 2 \times 0 + 1 \times 3 & 2 \times (-1) + 1 \times 2 \\ -3 \times 0 + 0 \times 3 & -3 \times (-1) + 0 \times 2 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 0 + 3 & -2 + 2 \\ 0 + 0 & 3 + 0 \end{bmatrix}$

$= \frac{1}{3} \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

• Postmultiplying $\text{A}$ with $\text{A}^{-1}$:

$\text{A}^{-1} \times \text{A} = \frac{1}{3} \begin{bmatrix} 0 & -1 \\ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix}$

$= \frac{1}{3} \begin{bmatrix} 0 \times 2 + (-1) \times (-3) & 0 \times 1 – 1 \times 0 \\ 3 \times 2 + 2 \times (-3) & 3 \times 1 + 2 \times 0 \end{bmatrix}$

$= \frac{1}{3} \begin{bmatrix} 0 + 3 & 0 – 0 \\ 6 – 6 & 3 + 0 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Note: Matrix multiplication is not commutative.

### Examples of Inverse of a Matrix

Example 1: Check whether $\text{A} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ and $\text{B} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$ are inverse of each other.

$\text{A} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ and $\text{B} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

The product of the two given matrices = $\text{AB} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

$= \begin{bmatrix} 2 \times 2 + 3 \times (-1) & 2 \times (-3) + 3 \times 2 \\ 1 \times 2 + 2 \times (-1) & 1 \times (-3) + 2 \times 2 \end{bmatrix}$

$= \begin{bmatrix} 4 – 3 & -6 + 6 \\ 2 – 2 & -3 + 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \text{I}$

Since, $\text{AB} = \text{I}$, therefore, $\text{A}$ and $\text{B}$ are inverse of each other.

Example 2: Check whether $\text{A} = \begin{bmatrix} 1 & -1 & 2 \\ 4 & 0 & 6 \\ 0 & 1 & -1 \end{bmatrix}$ and $\text{B} = \begin{bmatrix} 3 & -\frac{1}{2} & 3 \\ -2 & \frac{1}{2} & -1 \\ -2 & \frac{1}{2} & -2 \end{bmatrix}$ are inverse of each other.

$\text{A} = \begin{bmatrix} 1 & -1 & 2 \\ 4 & 0 & 6 \\ 0 & 1 & -1 \end{bmatrix}$ and $\text{B} = \begin{bmatrix} 3 & -\frac{1}{2} & 3 \\ -2 & \frac{1}{2} & -1 \\ -2 & \frac{1}{2} & -2 \end{bmatrix}$

The product of the two given matrices = $\text{AB} = \begin{bmatrix} 1 & -1 & 2 \\ 4 & 0 & 6 \\ 0 & 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -\frac{1}{2} & 3 \\ -2 & \frac{1}{2} & -1 \\ -2 & \frac{1}{2} & -2 \end{bmatrix}$

$= \begin{bmatrix} 1 \times 3 – 1 \times (-2) + 2 \times (-2) & 1 \times \left(-\frac{1}{2} \right) – 1 \times \frac{1}{2} + 2 \times \frac{1}{2} & 1 \times 3 – 1 \times (-1) + 2 \times (-2) \\ 4 \times 3 + 0 \times (-2) + 6 \times (-2) & 4 \times \left(-\frac{1}{2} \right) + 0 \times \frac{1}{2} + 6 \times \frac{1}{2} & 4 \times 3 + 0 \times (-1) + 6 \times (2) \\ 0 \times 3 + 1 \times (-2) – 1 \times (-2) & 0 \times \left(-\frac{1}{2} \right) + 1 \times \frac{1}{2} – 1 \times \frac{1}{2} & 0 \times 3 + 1 \times (-1) – 1 \times (-2) \end{bmatrix}$

$= \begin{bmatrix} 3 + 2 – 4 & -\frac{1}{2} – \frac{1}{2} + 1 & 3 + 1 – 4 \\ 12 + 0 – 12 & -2 + 0 + 3 & 12 + 0 – 12 \\ 0 – 2+ 2 & 0 + \frac{1}{2} – \frac{1}{2} & 0 – 1 + 2 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \text{I}$

Since, $\text{AB} = \text{I}$, therefore, $\text{A}$ and $\text{B}$ are inverse of each other.

## How to Find Inverse of a Matrix?

One of the most important methods of finding the matrix inverse involves finding the minors and cofactors of elements of the given matrix. The steps to find an inverse of a matrix are as follows.

• The inverse matrix is also found using the following equation $\text{A}^{-1} = \frac{\text{adj}(\text{A})}{\text{det}(\text{A})}$,  where $\text(adj)(\text{A})$ refers to the adjoint of a matrix $\text{A}$, $\text{det}( \text{A})$ refers to the determinant of a matrix $\text{A}$.
• The adjoint of a matrix $\text{A}$ or $\text{adj}( \text{A})$ can be found using the following method.
• In order to find the adjoint of a matrix $\text{A}$ first, find the cofactor matrix of a given matrix and then take the transpose of a cofactor matrix.
• The cofactor of a matrix can be obtained as $\text{C}_{ij} = (-1)^{i+j} \text{det} \left(\text{M}_{ij} \right)$

Here, $\text{M}_{ij}$ refers to the $(i,j)^{th}$ minor matrix after removing the $i^{th}$ row and the $j^{th}$ column. The transpose of a cofactor matrix is also called the adjoint of a matrix $\text{A}$.

### Examples of How to Find Inverse of a Matrix

Example 1: Find inverse of the matrix $\text{A} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$.

$\text{A} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$

$|\text{A}| = 2 \times 2 – 3 \times 1 = 4 – 3 = 1$

Now, we will find the minors of the matrix $\text{A}$.

$\text{M}_{1, 1} = 2$

$\text{M}_{1, 2} = 1$

$\text{M}_{2, 1} = 3$

$\text{M}_{2, 2} = 2$

Next, we will find the cofactors of the matrix $\text{A}$.

$\text{C}_{1, 1} = (-1)^{1 + 1} \text{M}_{1, 1} = 1 \times 2 = 2$

$\text{C}_{1, 2} = (-1)^{1 + 2} \text{M}_{1, 2} = -1 \times 1 = -1$

$\text{C}_{2, 1} = (-1)^{2 + 1} \text{M}_{2, 1} = -1 \times 3 = -3$

$\text{C}_{2, 2} = (-1)^{2 + 2} \text{M}_{2, 2} = 1 \times 2 = 2$

Therefore, cofactors matrix is $\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$

Now, we will find the transpose of the cofactors matrix, which is $\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

Therefore, $\text{Adj}(\text{A}) = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

Thus $\text{A}^{-1} = \frac{\text{Adj}(A)}{|\text{A}|} = \frac{\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}}{1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$.

Example 2: Find inverse of the matrix $\text{A} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 2 & 9 \end{bmatrix}$.

$\text{A} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 2 & 9 \end{bmatrix}$

$|\text{A}| = 1 \times \begin{bmatrix} 5 & 6 \\ 2 & 9 \end{bmatrix} – 2 \times \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix} + 3 \times \begin{bmatrix} 4 & 5 \\ 7 & 2 \end{bmatrix}$

$= 1 \times (5 \times 9 – 6 \times 2) – 2 \times (4 \times 9 – 6 \times 7) + 3 \times (4 \times 2 – 5 \times 7)$

$= 1 \times (45 – 12) – 2 \times (36 – 42) + 3 \times (8 – 35) = 1 \times 33 – 2 \times (-6) + 3 \times (-27)$

$= 33 + 12 – 81 = -36$

Now, we will find the minors of the matrix $\text{A}$.

$\text{M}_{1, 1} = \begin{bmatrix} 5 & 6 \\ 2 & 9 \end{bmatrix} = 5 \times 9 – 6 \times 2 = 45 – 12 = 33$

$\text{M}_{1, 2} = \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix} = 4 \times 9 – 6 \times 7 = 36 – 42 = -6$

$\text{M}_{1, 3} = \begin{bmatrix} 4 & 5 \\ 7 & 2 \end{bmatrix} = 4 \times 2 – 5 \times 7 = 8 – 35 = -27$

$\text{M}_{2, 1} = \begin{bmatrix} 2 & 3 \\ 2 & 9 \end{bmatrix} = 2 \times 9 – 3 \times 2 = 18 – 6 = 12$

$\text{M}_{2, 2} = \begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix} = 1 \times 9 – 3 \times 7 = 9 – 21 = -12$

$\text{M}_{2, 3} = \begin{bmatrix} 1 & 2 \\ 7 & 2 \end{bmatrix} = 1 \times 2 – 2 \times 7 = 2 – 14 = -12$

$\text{M}_{3, 1} = \begin{bmatrix} 2 & 3 \\ 5 & 6 \end{bmatrix} = 2 \times 6 – 3 \times 5 = 12 – 15 = -3$

$\text{M}_{3, 2} = \begin{bmatrix} 1 & 3 \\ 4 & 6 \end{bmatrix} = 1 \times 6 – 3 \times 4 = 6 – 12 = -6$

$\text{M}_{3, 3} = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} = 1 \times 5 – 2 \times 4 = 5 – 8 = -3$

Next, we will find the cofactors of the matrix $\text{A}$.

$\text{C}_{1, 1} = (-1)^{1 + 1} \text{M}_{1, 1} = 1 \times 33 = 33$

$\text{C}_{1, 2} = (-1)^{1 + 2} \text{M}_{1, 2} = -1 \times (-6) = 6$

$\text{C}_{1, 3} = (-1)^{1 + 3} \text{M}_{1, 3} = 1 \times (-27) = -27$

$\text{C}_{2, 1} = (-1)^{2 + 1} \text{M}_{2, 1} = -1 \times 12 = -12$

$\text{C}_{2, 2} = (-1)^{2 + 2} \text{M}_{2, 2} = 1 \times (-12) = -12$

$\text{C}_{2, 3} = (-1)^{2 + 3} \text{M}_{2, 3} = -1 \times (-12) = 12$

$\text{C}_{3, 1} = (-1)^{3 + 1} \text{M}_{3, 1} = 1 \times (-3) = -3$

$\text{C}_{3, 2} = (-1)^{3 + 2} \text{M}_{3, 2} = -1 \times (-6) = 6$

$\text{C}_{3, 3} = (-1)^{3 + 3} \text{M}_{3, 3} = 1 \times (-3) = -3$

Therefore, cofactors matrix is $\begin{bmatrix} 33 & 6 & -27 \\ -12 & -12 & 12 \\ -3 & 6 & -3 \end{bmatrix}$

Now, we will find the transpose of the cofactors matrix, which is $\begin{bmatrix} 33 & -12 & -3 \\ 6 & -12 & 6 \\ -27 & 12 & -3 \end{bmatrix}$

Therefore, $\text{Adj}(\text{A}) = \begin{bmatrix} 33 & -12 & -3 \\ 6 & -12 & 6 \\ -27 & 12 & -3 \end{bmatrix}$

Thus $\text{A}^{-1} = \frac{\text{Adj}(A)}{|\text{A}|} = \frac{\begin{bmatrix} 33 & -12 & -3 \\ 6 & -12 & 6 \\ -27 & 12 & -3 \end{bmatrix}}{-36}$

$= \begin{bmatrix} -\frac{11}{12} & \frac{1}{3} & \frac{1}{12} \\ -\frac{1}{6} & \frac{1}{3} & -\frac{1}{6} \\ \frac{3}{4} & -\frac{1}{3} & \frac{1}{12} \end{bmatrix}$

## Properties of Inverse Matrix

The following are the properties of the inverse matrix:

• If $\text{A}$ and $\text{B}$  are square matrices with the order $n$  and their product is an identity matrix, i.e., $\text{AB} = \text{I}_n = \text{BA}$, then $\text{B} = \text{A}^{–1}$.
• If a square matrix $\text{A}$  has an inverse (non-singular), then the inverse matrix is unique.
• A square matrix $\text{A}$  has an inverse matrix if and only if the determinant is not zero, i.e., $| \text{A}| \ne 0$. Similarly, the matrix $\text{A}$ is singular (has no inverse) if and only if its determinant is zero, i.e., $| \text{A}| = 0$.
• If a square matrix $\text{A}$ has an inverse, then the determinant of an inverse matrix is the reciprocal of the matrix determinant. i.e., $∣\text{A}^{–1}∣ = \frac{1}{| \text{A}|}$.
• If a square matrix $\text{A}$  has an inverse, for a scalar $k \ne 0$  then the inverse of a scalar multiple is equal to the product of their inverse, i.e., $(k \text{A})^{–1} = 1k\text{A}^{–1}$.
• If a square matrix $\text{A}$ has an inverse, then the transpose of an inverse matrix is equal to the inverse of the transposed matrix. i.e., $\left( \text{A}^{–1} \right)^{\text{T}} = \left( \text{A}^{\text{T}} \right)^{–1}$.
• If $\text{A}$ and $\text{B}$  are square non-singular matrices both with the order n then the inverse of their product is equal to the product of their inverse in reverse order, i.e., $( \text{AB})^{–1} = \text{B}^{–1} \text{A}^{–1}$.
• Let $\text{A}$ and $\text{B}$ be square matrices with the order n. If $\text{AB} = \text{O}$  then either $\text{A} = \text{O}$  or $\text{B} = \text{O}$  or both $\text{A}$  and $\text{B}$  are singular matrices with no inverse.

## Practice Problems

Find the inverse of the following matrices

• $\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}$
• $\begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix}$
• $\begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{bmatrix}$
• $\begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{bmatrix}$
• $\begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & -1 \\ 2 & 3 & 2 \end{bmatrix}$

## FAQs

### What is the meaning of inverse of matrix?

The inverse of a matrix is another matrix, which on multiplication with the given matrix gives the multiplicative identity. For a matrix, $\text{A}$ its inverse is denoted by $\text{A}^{-1}$, and $\text{A} \times \text{A}^{-1} = \text{A}^{-1} \times \text{A} = \text{I}$, where $\text{I}$ is the identity matrix.

### What is invertible matrix?

The matrix whose determinant is non-zero and for which the inverse matrix can be calculated is called an invertible matrix. For example, the inverse of $\text{A} = \begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix}$ is $\text{A}^{-1} = \frac{1}{3} \begin{bmatrix} 0 & -1 \\ 3 & 2 \end{bmatrix}$.

### Is the sum of a matrix and its inverse a zero matrix?

No, the sum of a matrix and its inverse is not a zero matrix. The product of a matrix with its inverse is a unit matrix.

### Is the product of a matrix with its inverse a zero matrix?

No, the product of a matrix with its inverse is not a zero matrix. The product of a matrix with its inverse is a unit matrix.

### How to find inverse of matrix?

The steps to find an inverse of a matrix are as follows.
a) The inverse matrix is also found using the following equation $\text{A}^{-1} = \frac{\text{adj}(\text{A})}{\text{det}(\text{A})}$,  where $\text(adj)(\text{A})$ refers to the adjoint of a matrix $\text{A}$, $\text{det}( \text{A})$ refers to the determinant of a matrix $\text{A}$.
b) The adjoint of a matrix $\text{A}$ or $\text{adj}( \text{A})$ can be found using the following method.
c) In order to find the adjoint of a matrix $\text{A}$ first, find the cofactor matrix of a given matrix and then take the transpose of a cofactor matrix.
d) The cofactor of a matrix can be obtained as $\text{C}_{ij} = (-1)^{i+j} \text{det} \left(\text{M}_{ij} \right)$

### What is the use of inverse of a matrix?

The inverse of a matrix is used to solve a system of linear equations.

### What is the formula for an inverse matrix?

The inverse of a matrix $\text{A}$ is evaluated by using the formula $\text{A}^{-1} = \frac{\text{adj}(\text{A})}{\text{det}(\text{A})}$,  where $\text(adj)(\text{A})$ refers to the adjoint of a matrix $\text{A}$, $\text{det}( \text{A})$ refers to the determinant of a matrix $\text{A}$.

## Conclusion

The inverse of a matrix is another matrix, which on multiplication with the given matrix gives the multiplicative identity. For a matrix, $\text{A}$ its inverse is denoted by $\text{A}^{-1}$, and $\text{A} \times \text{A}^{-1} = \text{A}^{-1} \times \text{A} = \text{I}$, where $\text{I}$ is the identity matrix. The inverse of a matrix is used to solve a system of linear equations.