- Written By
Ritesh Kumar Gupta
- Last Modified 25-01-2023

# Inverse Matrix: Meaning, Formula, Solved Examples

**Inverse Matrix:** A matrix is a set of objects that are arranged in rows and columns. These items are known as matrix elements. A matrix’s order is stated as the number of rows divided by the number of columns. For example, 2 × 2, 2 × 3, 3 × 2, 3 × 3, 4 × 4 and so on. Only square matrices with the same number of rows and columns can have their inverse determined. Inverse Matrix is an important tool in the mathematical world.

It is used in solving a system of linear equations. Inverse matrices are frequently used to encrypt or decrypt message codes. It is also used to explore electrical circuits, quantum mechanics, and optics. These matrices are crucial in measuring battery power outputs and converting electrical energy into other useable energy by resistors. When applying Kirchhoff’s laws of voltage and current to solve problems, the inverse matrices are extremely significant. Continue reading to know more.

## Inverse Matrix Definition

Before learning about the inverse matrix, we must know what is an adjoint of a matrix.

The adjoint of a square matrix \(A = {\left[ {{a_{ij}}} \right]_{n \times n}}\) is the transpose of the matrix \({\left[ {{A_{ij}}} \right]_{n \times n}}\) where \({A_{ij}}\) is the cofactor of the element \({a_{ij}}\) The adjoint of the matrix \(A\) is denoted by \(Adj{\rm{ }}A .\)

Let,

\(A = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\
{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\
{{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]\)

Then \(Adj{\rm{ }}A = \) Transpose of \(\left[ {\begin{array}{*{20}{c}}
{{A_{11}}}&{{A_{12}}}&{{A_{13}}}\\
{{A_{21}}}&{{A_{22}}}&{{A_{23}}}\\
{{A_{31}}}&{{A_{32}}}&{{A_{33}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\
{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\
{{A_{13}}}&{{A_{23}}}&{{A_{33}}}
\end{array}} \right]\)

where \({A_{ij}}\) is cofactor of \({a_{ij}}\) which is calculated by using the relation \({A_{ij}} = {( – 1)^{i + j}}{M_{ij}}\), where \({M_{ij}}\) is minor of \({a_{ij}}\).

Minor of an element \({a_{ij}}\) of a determinant is the determinant obtained by removing its i th row and j th column in which element \({a_{ij}}\) lies. Minor of an element \({a_{ij}}\) is denoted by \({M_{ij}}\).

In mathematics, a cofactor is used to find the inverse and adjoint of a matrix. If

\(A = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\
{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\
{{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]\)

then \(adj A = {C^T}\), where \(C\) is the cofactor matrix of \(A\).

Learn About Matrices Here

See the below diagram to know how the adjoint of a matrix \(A\) is calculated.

### How To Calculate Inverse Matrix?

If you are asked what is the inverse of a number \(5\)?, you would quickly say \(\frac{1}{5}\). But have you ever thought about why it is the inverse of \(5\)? It is because \(5 \times \frac{1}{5} = 1\) which is a multiplicative identity.

Similar is the case with the matrices; if we are asked about the inverse of a matrix \(A\), then what should be multiplied with \(A\) to get an identity matrix that is basically the inverse of matrix \(A\).

Hence, if \(A\) is a square matrix of order \(n\), and if there exists another square matrix \(B\) of the same order \(n\), such that \(AB=BA=I\), then \(B\) is called the inverse matrix of \(A\) and it is denoted by \({A^{ – 1}}\).

Since, \(A(adjA) = (adjA)A = |A|I \Rightarrow \frac{{A(adjA)}}{{|A|}} = \frac{{(adjA)}}{{|A|}} = I\)

So, by comparing with the above definition, \({A^{ – 1}} = \frac{{adjA}}{{|A|}}\) where \(|A| \ne 0\)

The following are the two methods to find the inverse of a matrix:

1. **Elementary Operations**

Suppose \(X, A\) and \(B\) be matrices of the same order such that \(X = AB\). In order to have a sequence of elementary row operations on the matrix equation \(X = AB\), we will apply these row operations simultaneously on \(X\) and on the first matrix \(A\) of the product \(AB\) on RHS. Similarly, in order to have a sequence of elementary column operations on the matrix equation \(X = AB\), we will apply these operations simultaneously on X and on the second matrix \(B\) of the product \(AB\) on \(RHS\).

In view of the above discussion, we conclude that if \(A\) is a matrix such that \({A^{ – 1}}\) exists, then to find \({A^{ – 1}}\) using elementary row operations, write \(A = IA\) and apply a sequence of row operation on \(A = IA\) till we get, \(I = BA\). The matrix \(B\) will be the inverse of A. Similarly, if we wish to find \({A^{ – 1}}\) using column operations, then, write \(A = AI\) and apply a sequence of column operations on \(A = AI\) till we get, \(I = AB\).

2. **Inverse Matrix Method or Matrix Inversion Method or Adjoint Method**

This method is used to solve the system of linear equations.

Let’s have a matrix equation:

\(AX=AB\)

Pre-multiply with \({A^{ – 1}}\) to both the sides

\( \Rightarrow {A^{ – 1}}AX = {A^{ – 1}}B\)

\( \Rightarrow IX = {A^{ – 1}}B\)

\( \Rightarrow X = {A^{ – 1}}B\)

\({A^{ – 1}}\) is calculated from the below relation

\({A^{ – 1}} = \frac{{adjA}}{{|A|}}\)

### Properties of Inverse Matrix

The following are the properties of the inverse matrix:

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

### Inverse Matrix: Examples

Suppose we want to calculate the inverse of a matrix \(A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
3&5&0\\
2&1&8
\end{array}} \right]\)

To find the inverse of a matrix, we first need to find the adjoint of matrix A.

Cofactor of \(1 = {A_{11}} = + \left| {\begin{array}{*{20}{c}}
5&0\\
1&8
\end{array}} \right| = + (40 – 0) = 40\)

Cofactor of \(2 = {A_{12}} = – \left| {\begin{array}{*{20}{c}}
3&0\\
2&8
\end{array}} \right| = – (24 – 0) = – 24\)

Cofactor of \(3 = {A_{13}} = + \left| {\begin{array}{*{20}{c}}
3&5\\
2&1
\end{array}} \right| = + (3 – 10) = – 7\)

Cofactor of \(4 = {A_{21}} = – \left| {\begin{array}{*{20}{c}}
0&0\\
1&8
\end{array}} \right| = 0\)

Cofactor of \(5 = {A_{22}} = + \left| {\begin{array}{*{20}{c}}
1&0\\
2&8
\end{array}} \right|= + (8 – 0) = 8\)

Cofactor of \(6 = {A_{23}} = – \left| {\begin{array}{*{20}{c}}
1&0\\
2&1
\end{array}} \right| = – (1 – 0) = – 1\)

Cofactor of \(7 = {A_{31}} = + \left| {\begin{array}{*{20}{c}}
0&0\\
5&0
\end{array}} \right| = 0\)

Cofactor of \(8 = {A_{32}} = – \left| {\begin{array}{*{20}{c}}
1&0\\
3&0
\end{array}} \right| = – (1 – 0) = – 1\)

Cofactor of \(9 = {A_{33}} = + \left| {\begin{array}{*{20}{c}}
1&0\\
3&5
\end{array}} \right| = + (5 – 0) = 5\)

The Cofactor matrix of A is

\(\left[ {{A_{ij}}} \right] = \left[ {\begin{array}{*{20}{c}}
{{A_{11}}}&{{A_{12}}}&{{A_{13}}}\\
{{A_{21}}}&{{A_{22}}}&{{A_{23}}}\\
{{A_{31}}}&{{A_{32}}}&{{A_{33}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{40}&{ – 24}&{ – 7}\\
0&8&{ – 1}\\
0&{ – 1}&5
\end{array}} \right]\)

Now find the transpose of \({A_{ij}}\)

\(adj\,A = {\left[ {\begin{array}{*{20}{c}}
{40}&{ – 24}&{ – 7}\\
0&8&{ – 1}\\
0&{ – 1}&5
\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}
{40}&0&0\\
{ – 24}&8&{ – 1}\\
{ – 7}&{ – 1}&5
\end{array}} \right]\)

Since \(|A| = 40\)

So, The inverse of matrix A will be \({A^{ – 1}} = \frac{{adj\,A}}{{\left| A \right|}} = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}}
{40}&0&0\\
{ – 24}&8&{ – 1}\\
{ – 7}&{ – 1}&5
\end{array}} \right]\)

### Applications of Inverse Matrix

The application of inverse matrix is as follows:

- The inverse matrix is used to solve the system of linear equations. It also tells us the consistent or inconsistent behaviour of the solution of equations.
- Linear equations, some error-correcting codes (linear codes), linear differential equations, and linear recurrence sequences all use the concept of the inverse matrix.
- Inverse matrices are frequently used to encrypt message codes. Matrices are used by programmers to code or encrypt letters. A message is made up of a series of binary numbers that are solved using coding theory for communication. As a result, the concept of matrices is used to solve such equations.
- Engineers and physicists develop models of physical structures and execute the precise calculations required to operate complicated machinery. Fine-tuned matrix transformation computations are used in electronics, networks, aeroplanes and spacecraft, and chemical processing.
- In physics, the Inverse matrix is used to explore electrical circuits, quantum mechanics, and optics. These matrices are crucial in the measuring of battery power outputs and the conversion of electrical energy into other useable energy by resistors. When applying Kirchhoff’s laws of voltage and current to solve problems, the inverse matrices are extremely significant.

### Solved Examples on Inverse Matrix

**Q.1. For the matrix \(A = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 4&5&6\\ 1&2&3 \end{array}} \right],\) Show that \({A^{ – 1}}\) will not exist.**

**Ans:** Given matrix is \(A = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 4&5&6\\ 1&2&3 \end{array}} \right]\)

Since, \(|A| = 0\) as two rows are the same \( \Rightarrow A\) is a singular matrix \( \Rightarrow {A^{ – 1}}\) will not exist.

**Q.2. Find the inverse of a matrix \(\left[ {\begin{array}{*{20}{c}} 7&1\\ 4&{ – 3} \end{array}} \right]\) using elementary operations**

**Ans: **Given that \(A = \left[ {\begin{array}{*{20}{c}} 7&1\\ 4&{ – 3} \end{array}} \right]\)

We know \(A = IA \Rightarrow \left[ {\begin{array}{*{20}{c}} 7&1\\ 4&{ – 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]A\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{\frac{1}{7}}\\ 4&{ – 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{7}}&0\\ 0&1 \end{array}} \right] A\) (Applying \({R_1} \to \frac{1}{7}{R_1}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{\frac{1}{7}}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{7}}&0\\ {\frac{4}{7}}&{ – \frac{7}{{25}}} \end{array}} \right]A\) (Applying \({R_2} \to {R_2} – 4{R_1}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{\frac{1}{7}}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{7}}&0\\ {\frac{4}{{25}}}&{ – \frac{7}{{25}}} \end{array}} \right]A\) (Applying \({R_2} \to – \frac{7}{{25}}{R_2}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{3}{{25}}}&{\frac{1}{{25}}}\\ {\frac{4}{{25}}}&{ – \frac{7}{{25}}} \end{array}} \right]A\) (Applying \({R_1} \to {R_1} – \frac{1}{7}{R_2}\))

\( \Rightarrow {A^{ – 1}} = \frac{1}{{25}}\left[ {\begin{array}{*{20}{c}} 3&1\\ 4&{ – 7} \end{array}} \right]\)

**Q.3. Find the inverse of a matrix \(\left[ {\begin{array}{*{20}{c}} 0&1&2\\ 1&2&3\\ 3&1&1 \end{array}} \right]\) using elementary operations**

Ans:

Given that \(A = \left[ {\begin{array}{*{20}{c}} 0&1&2\\ 1&2&3\\ 3&1&1 \end{array}} \right]\)

We know \(A = IA \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&1&2\\ 1&2&3\\ 3&1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]A\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&1&2\\ { – 2}&1&2\\ 3&1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&{ – 1}\\ 0&0&1 \end{array}} \right]A\) (Applying \({R_2} \to {R_2} – {R_3}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&1&2\\ { – 2}&1&2\\ 3&1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&{ – 1}\\ 0&0&1 \end{array}} \right]A\) (Applying \({R_1} \to {R_1} – {R_3}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} { – 3}&0&1\\ 0&3&4\\ 0&{ – 2}&{ – 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&{ – 1}\\ { – 2}&3&{ – 1}\\ 3&{ – 3}&1 \end{array}} \right]A\) (Applying \({R_2} \to 3{R_2} – 2{R_1}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} { – 3}&0&1\\ 0&3&4\\ 0&{ – 2}&{ – 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&{ – 1}\\ { – 2}&3&{ – 1}\\ 3&{ – 3}&1 \end{array}} \right]A\) (Applying \({R_3} \to {R_3} – {R_2}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} { – 3}&0&1\\ 0&3&4\\ 0&1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&{ – 1}\\ { – 2}&3&{ – 1}\\ {\frac{{ – 3}}{2}}&{\frac{3}{2}}&{\frac{{ – 1}}{2}} \end{array}} \right]A\) (Applying \({R_2} \to {R_2} – 4{R_3}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} { – 3}&0&1\\ 0&{ – 1}&0\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&{ – 1}\\ 4&{ – 3}&1\\ {\frac{5}{2}}&{\frac{{ – 3}}{2}}&{\frac{1}{2}} \end{array}} \right]A\) (Applying \({R_3} \to {R_3} – {R_2}\))

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{{ – 1}}{2}}&{\frac{1}{2}}\\ { – 4}&3&{ – 1}\\ {\frac{5}{2}}&{\frac{{ – 3}}{2}}&{\frac{1}{2}} \end{array}} \right]A\) (Applying \({R_2} \to – {R_2}\))

\( \Rightarrow {A^{ – 1}} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{{ – 1}}{2}}&{\frac{1}{2}}\\ { – 4}&3&{ – 1}\\ {\frac{5}{2}}&{\frac{{ – 3}}{2}}&{\frac{1}{2}} \end{array}} \right]\)

**Q.4. Solve the following system of linear equations using matrix inversion method: \(5x + 2y = 3,\,3x + 2y = 5\)**

Ans:

The matrix form of the system is \(AX = B,\) where \(A = \left[ {\begin{array}{*{20}{c}} 5&2\\ 3&2 \end{array}} \right],\,X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right],\,B = \left[ {\begin{array}{*{20}{c}} 3\\ 5 \end{array}} \right].\)

We find \(\left| A \right| = \left| {\begin{array}{*{20}{c}} 5&2\\ 3&2 \end{array}} \right| = 10 – 6 = 4 \ne 0.\) So, \({A^{ – 1}}\) exists and \({A^{ – 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 2&{ – 2}\\ { – 3}&5 \end{array}} \right]\)

Then, from \(X = {A^{ – 1}}B,\) we get

\(\left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 2&{ – 2}\\ { – 3}&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\\ 5 \end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 6&{ – 10}\\ { – 9}&{ + 25} \end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} { – 4}\\ {16} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{ – 4}}{4}}\\ {\frac{{16}}{4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { – 1}\\ 4 \end{array}} \right]\)

Hence, the solution is \(\left( {x = – 1,\,y = 4} \right).\)

**Q.5. Solve the following system of equations using the matrix inversion method: \(2{x_1} + 3{x_2} + 3{x_3} = 5**

{x_1} – 2{x_2} + {x_3} = – 4

3{x_1} – {x_2} – 2{x_3} = 3\)

Ans: The matrix form of the system is \(AX=B\), where

\(A = \left[ {\begin{array}{*{20}{c}} 2&3&3\\ 1&{ – 2}&2\\ 3&{ – 1}&{ – 2} \end{array}} \right],\,X = \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right],\,B = \left[ {\begin{array}{*{20}{c}} 5\\ { – 4}\\ 3 \end{array}} \right].\) We find \(\left| A \right| = \left[ {\begin{array}{*{20}{c}} 2&3&3\\ 1&{ – 2}&2\\ 3&{ – 1}&{ – 2} \end{array}} \right] = 2\left( {4 + 1} \right) – 3\left( { – 2 – 3} \right) + 3\left( { – 1 + 6} \right) = 10 + 15 + 15 = 40 \ne 0.\)

So, \({A^{ – 1}}\) exists and

\({A^{ – 1}} = \frac{1}{{\left| A \right|}}\left( {adj\,A} \right) = \frac{1}{{40}}{\left[ {\begin{array}{*{20}{c}} { + \left( {4 + 1} \right)}&{ – \left( { – 2 – 3} \right)}&{ + \left( { – 1 + 6} \right)}\\ { – \left( { – 6 + 3} \right)}&{ + \left( { – 4 – 9} \right)}&{ – \left( { – 2 – 9} \right)}\\ { + \left( {3 + 6} \right)}&{ – \left( {2 – 3} \right)}&{ + \left( { – 4 – 3} \right)} \end{array}} \right]^T} = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} 5&3&9\\ 5&{ – 13}&1\\ 5&{11}&{ – 7} \end{array}} \right]\)

Then, applying \(X = {A^{ – 1}}B,\) we get \(\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} 5&3&9\\ 5&{ – 13}&1\\ 5&{11}&{ – 7} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5\\ { – 4}\\ 3 \end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} {25}&{ – 12}&{ + 27}\\ {25}&{ + 52}&{ + 3}\\ {25}&{ – 44}&{ – 21} \end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} {40}\\ {80}\\ { – 40} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ 2\\ { – 1} \end{array}} \right]\)

Hence, the solution is \(\left( {{x_1} = 1,{x_2} = 2,{x_3} = – 1} \right)\)

### Summary

An inverse matrix is an important tool in mathematics. We have learnt about the inverse matrix, its properties, and its examples. It can be used to solve the bulk of difficult problems. It is used for solving linear equations and other mathematical functions such as calculus, optics, and quantum physics. It has a wide range of real-world applications, which has led to it playing a crucial role in mathematics.

### FAQs on Inverse Matrix

The following are the most frequently asked questions on Inverse Matrix:

**Q.1: What is the inverse of \(2 \times 2\) matrix?**

Ans: For a square matrix of order \(2\) , given by \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)

The \(adj A\) can be obtained by interchanging \({{a_{11}}}\) and \({{a_{22}}}\) and by changing signs of \({{a_{12}}}\) and \({{a_{21}}}\), i.e.,

\(adj\,A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{ – {a_{12}}}\\ { – {a_{21}}}&{{a_{22}}} \end{array}} \right]\)

Then,

\({A^{ – 1}} = \frac{{adjA}}{{|A|}}\)

**Q.2: What is matrix inversion method?**

Ans: This method is used to solve the system of linear equations. Consider a matrix equation \(AX=B\)

Pre-multiply with \({A^{ – 1}}\) to the both sides

\( \Rightarrow {A^{ – 1}}AX = {A^{ – 1}}B\)

\( \Rightarrow IX = {A^{ – 1}}B\)

\( \Rightarrow X = {A^{ – 1}}B\)

\({A^{ – 1}}\) is calculated from the below relation

\({A^{ – 1}} = \frac{{adjA}}{{|A|}}\)

**Q.3: How do you find the inverse of a \(3 \times 3\) matrix?**

Ans: If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\)

Then \(adj (A) = {C^T},\) where C is the cofactor matrix of \(A.\)

**Q.4: Can a matrix have \(2\) inverse?**

Ans: No, a matrix cannot have \(2\) inverse. The inverse of a matrix is unique. This will be proved with the help of the contradiction method.

**Q.5: What is the use of inverse matrix?**

Ans: Inverse matrix is used to solve the system of linear equations. It is frequently used to encrypt message codes. Matrices are used by programmers to code or encrypt letters. A message is made up of a series of binary numbers that are solved using coding theory for communication and then an inverse matrix is used to decrypt the encoded message.

Learn About Uses of Matrices Here

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