Adjoint of a matrix: It is the simplest method for calculating a matrix’s inverse. A matrix is an ordered rectangular array of numbers or functions in linear algebra. The numbers or functions are referred to as the matrix’s elements or entries. Also, matrixes can be classified according to the number of rows and columns in which items are placed.

An adjoint matrix is also known as an adjugate matrix. It is used in areas of business and science like budgeting, sales projection, and cost estimation. It has a role in other branches like genetics, economics, sociology and industrial management. Let us learn more about the properties of the adjoint of a matrix 2×2 and 3X3, how to find the adjoint of different matrices, adjoint of a matrix formula and examples.

What is the Adjoint of a Matrix?

Before learning about the adjoint of a matrix, we must know what is a matrix first? A matrix (plural matrices) is a rectangular table or array containing numbers, symbols, or expressions organised in rows and columns.

A matrix is generally represented by a capital letter in a boldface font (e.g., \(A, B, X\)) and the elements of the matrix are represented by lowercase letters with a double subscript (e.g., \(a_{ij},\,b_{ij},\,x_{ij}\)).
For instance: In the matrix \(A\), \(a_{23}\) is an element in the second row and the third column. A \(3 × 3\) matrix \(A\) is shown below

Adjoint of a Matrix Formula

You might be wondering how to find the adjoint of a matrix? By using the cofactor and transpose of a matrix, the formula for the adjoint of a matrix may be derived. The adjugate matrix for a 2 x 2 matrix, on the other hand, is simple to find. Let’s have a look at the formulas and steps involved in determining the adjoint matrix for a given matrix.

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

A general representation of matrix \(A\) of \(n × m\) order is given below.
Here, \(n\) represents the number of rows in \(A\)
And \(m\) represent the number of columns in \(A\)

In brief, the above matrix is represented by \(A = {\left[ {{a_{ij}}} \right]_{\left( {n \times m} \right)}}.\) The number \(a_{11}\), \(a_{12}\), …..etc., are known as the elements of the matrix \(A\), where, \(a_{ij}\) belongs to the \(i\)th row and \(j\)th column and is called the \((i, j)\)th element of the matrix \(A = [a_{ij}]\).
Let us now discuss the adjoint of a matrix.

The adjoint of a square matrix \(A=[a_{ij} ]_{(n×n)}\) is defined as the transpose of the matrix \([A_{ij}]_{(n×n)}\), where \(A_{ij}\) is the cofactor of the element \(a_{ij}\). The adjoint of the matrix \(A\) is denoted by \(adj\,A\).
Let \(A = \left[{\begin{array}{{c}} {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\ \end{array} } \right]\)

Then \(Adj\,A =\) Transpose of \(\left[{\begin{array}{{c}} {{A_{11}}} & {{A_{12}}} & {{A_{13}}} \\ {{A_{21}}} & {{A_{22}}} & {{A_{23}}} \\ {{A_{31}}} & {{A_{32}}} & {{A_{33}}} \\ \end{array} } \right] = \left[{\begin{array}{{c}} {{A_{11}}} & {{A_{21}}} & {{A_{31}}} \\ {{A_{12}}} & {{A_{22}}} & {{A_{32}}} \\ {{A_{13}}} & {{A_{23}}} & {{A_{33}}} \\ \end{array} } \right]\) 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 deleting 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 \(A = \left[{\begin{array}{{c}} {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\ \end{array} } \right]\)
then \(adj\left( A \right) = {C^T},\) where \(C\) is the cofactor matrix of \(A\).

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

Properties of Adjoint of a Matrix

The following are the properties of the adjoint of a matrix calculator which are very useful to solve a lot of mathematical problems:

1. \(A.adj⁡(A)=adj⁡(A)⋅A=|A|I\)
   where, \(A\) is a square matrix, \(I\) is an identity matrix of same order as of \(A\) and \(\left|A \right|\) represents determinant of matrix \(A\).
2. \(|adjA|=|A|^{n-1}\)
   Determinant of adjoint \(A\) is equal to determinant of A power \(n-1\) where \(A\) is invertible \(n \times n\) square matrix.
3. \(adj⁡(adj\,⁡A)=|A|^{n-2}⋅A\) where \(A\) is \(n \times n\) invertible square matrix.
4. \(adj⁡(AB)=adj⁡(B)⋅adj⁡(A)\)
5. \((adj⁡(A))^T=adj⁡(A^T )\)
6. \(adj⁡(kA) = k^{n-1} adj⁡(A)\)

How to find Adjoint of a Matrix?

Suppose we want to calculate the adjoint of a matrix \(A = \left[{\begin{array}{{c}} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{array} } \right]\)
To find the adjoint of a matrix, we first find the Cofactor matrix of \(A\) and then find the transpose of this Cofactor matrix \(A\).
Cofactor of \(1 = {A_{11}} = + \left| {\begin{array}{*{20}{c}} 5&6\\ 8&9 \end{array}} \right| = + \left( {45 – 48} \right) = – 3\)
Cofactor of \(2 = {A_{12}} = – \left| {\begin{array}{*{20}{c}} 4&6\\ 7&9 \end{array}} \right| = – \left( {36 – 42} \right) = 6\)
Cofactor of \(3 = {A_{13}} = + \left| {\begin{array}{*{20}{c}} 4&5\\ 7&8 \end{array}} \right| = + \left( {32 – 35} \right) = – 3\)
Cofactor of \(4 = {A_{21}} = – \left| {\begin{array}{*{20}{c}} 2&3\\ 8&9 \end{array}} \right| = – \left( {18 – 24} \right) = 6\)
Cofactor of \(5 = {A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&3\\ 7&9 \end{array}} \right| = + \left( {9 – 21} \right) = – 12\)
Cofactor of \(6 = {A_{23}} = – \left| {\begin{array}{*{20}{c}} 1&2\\ 7&8 \end{array}} \right| = – \left( {8 – 14} \right) = 6\)
Cofactor of \(7 = {A_{31}} = + \left| {\begin{array}{*{20}{c}} 2&3\\ 5&6 \end{array}} \right| = + \left( {12 – 15} \right) = – 3\)
Cofactor of \(8 = {A_{32}} = – \left| {\begin{array}{*{20}{c}} 1&3\\ 4&6 \end{array}} \right| = – \left( {6 – 12} \right) = 6\)
Cofactor of \(9 = {A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&2\\ 4&5 \end{array}} \right| = + \left( {5 – 8} \right) = – 3\)

The Cofactor matrix of \(A\) is \(\left[{{A_{ij}}} \right]=\left[{\begin{array}{{c}} {{A_{11}}} & {{A_{12}}} & {{A_{13}}} \\ {{A_{21}}} & {{A_{22}}} & {{A_{23}}} \\ {{A_{31}}} & {{A_{32}}} & {{A_{33}}} \\ \end{array} } \right] = \left[{\begin{array}{{c}} { – 3} & 6 & { – 3} \\ 6 & { – 12} & 6 \\ { – 3} & 6 & { – 3} \\ \end{array} } \right]\)

Now find the transpose of \(A_{ij}\)
\(adj\,A = {\left[{\begin{array}{{c}} { – 3} & 6 & { – 3} \\ 6 & { – 12} & 6 \\ { – 3} & 6 & { – 3} \\ \end{array} } \right]^T} = \left[{\begin{array}{{c}} { – 3} & 6 & { – 3} \\ 6 & { – 12} & 6 \\ { – 3} & 6 & { – 3} \\ \end{array} } \right]\)

Applications of Adjoint of a Matrix

The application or function of the adjoint of a matrix is as follows:

1. It is used to find the inverse of a matrix. The inverse of a matrix \(A\) is denoted as a matrix \(A^{-1}\) such that the multiplication of the given matrix \(A\) by \(A^{-1}\) is the identity matrix \(I\) \(A {A^ { – 1}} = I.\). An inverse matrix exists only for square non-singular matrices (whose determinant \(≠ 0\)). If \(A\) is a square non-singular matrix of order \(n\), the inverse matrix \(A^{-1}\) is given by \( {A^ { – 1}} = \frac{ {adj\,A}}{ {\left| A \right|}}\) Where \(adj\,A\) is the adjoint of the matrix and \(|A|\) is the determinant of the matrix \(A\).

2. The adjoint of a matrix formula helps solve the system of linear equations. It tells us whether the solution of equations is consistent or inconsistent.

3. Linear equations, some error-correcting codes (linear codes), linear differential equations, and linear recurrence sequences all use the matrix concept.

4. Matrices are frequently used to encrypt message codes. Programmers use matrixes and their inverse matrices to code or encrypt letters. A message comprises 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.

5. 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.

6. In physics, matrix algebra is 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. The matrices are extremely significant when applying Kirchhoff’s laws of voltage and current to solve problems.

7. In Google search, page ranking algorithms use Stochastic matrices and Eigenvectors to score websites.

8. Many IT organisations employ matrix data structures to track user information, run search queries, and maintain databases. In information security, many frameworks are built to work with matrices. Matrices are employed in electronic data compression, such as handling biometric data in Mauritius’ new Identity Card.

Solved Examples

Q.1: \(A = \left[{\begin{array}{{c}} 1 & { – 1} & 1 \\ 2 & 3 & 0 \\ {18} & 2 & {10} \\ \end{array} } \right]\). Show that \(A\left( {adj\,A} \right) = 0\)
Ans: Given matrix is \(A = \left[{\begin{array}{{c}} 1 & { – 1} & 1 \\ 2 & 3 & 0 \\ {18} & 2 & {10} \\ \end{array} } \right]\).
Now,
\({A_{11}} = \left| {\begin{array}{*{20}{c}} 3&0\\ 2&{10} \end{array}} \right| = 30,\,{A_{12}} = – \left| {\begin{array}{*{20}{c}} 2&0\\ {18}&{10} \end{array}} \right| = – 20,\,{A_{13}} = \left| {\begin{array}{*{20}{c}} 2&3\\ {18}&2 \end{array}} \right| = – 50\)
\({A_{21}} = – \left| {\begin{array}{*{20}{c}} { – 1}&1\\ 2&{10} \end{array}} \right| = 12,\,{A_{22}} = \left| {\begin{array}{*{20}{c}} 1&1\\ {18}&{10} \end{array}} \right| = – 8,\,{A_{23}} = – \left| {\begin{array}{*{20}{c}} 1&{ – 1}\\ {18}&2 \end{array}} \right| = – 20\)
\({A_{31}} = \left| {\begin{array}{*{20}{c}} { – 1}&1\\ 3&0 \end{array}} \right| = – 3,\,{A_{32}} = – \left| {\begin{array}{*{20}{c}} 1&1\\ 2&0 \end{array}} \right| = 2,\,{A_{33}} = \left| {\begin{array}{*{20}{c}} 1&{ – 1}\\ 2&3 \end{array}} \right| = 5\)
\(adj\,A = {\left[ {\begin{array}{*{20}{c}} {30}&{ – 20}&{ – 50}\\ {12}&{ – 8}&{ – 20}\\ { – 3}&2&5 \end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}} {30}&{12}&{ – 3}\\ { – 20}&{ – 8}&2\\ { – 50}&{ – 20}&5 \end{array}} \right]\)
\(\therefore \,A\,\left( {adj\,A} \right) = \left[ {\begin{array}{*{20}{c}} 1&{ – 1}&1\\ 2&3&0\\ {18}&2&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {30}&{12}&{ – 3}\\ { – 20}&{ – 8}&2\\ { – 50}&{ – 20}&5 \end{array}} \right]\)
\( = \left[ {\begin{array}{*{20}{c}} {30 + 20 – 50}&{12 + 8 – 20}&{ – 3 – 2 + 5}\\ {60 – 60 – 0}&{24 – 24 – 0}&{ – 6 + 6 + 0}\\ {540 – 40 – 500}&{216 – 16 – 200}&{ – 54 + 4 + 50} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&0\\ 0&0&0\\ 0&0&0 \end{array}} \right]\)

Q.2. If matrix \(A\) of order \(3\) has determinant value \(= 4\), then find the value of \(|adj\,A|\)
Ans: \(|adj\,A|=|A|^{n-1}= 4^2 = 16\)

Q.3: Compute the adjoint of the following matrix \(\left[{\begin{array}{{c}} 1 & 2 & 5 \\ 2 & 3 & 1 \\ { – 1} & 1 & 1 \\ \end{array} } \right]\). Verify that \(A.adj⁡(A) = adj⁡(A)⋅A = |A|I\)
Ans:
Given that \(A = \left[{\begin{array}{{c}} 1 & 2 & 5 \\ 2 & 3 & 1 \\ { – 1} & 1 & 1 \\ \end{array} } \right]\)
Now, \({A_{11}} = \left| {\begin{array}{*{20}{c}} 3&1\\ 1&1 \end{array}} \right| = 2,\,{A_{12}} = – \left| {\begin{array}{*{20}{c}} 2&1\\ { – 1}&1 \end{array}} \right| = – 3\) and \({A_{13}} = \left| {\begin{array}{*{20}{c}} 2&3\\ { – 1}&1 \end{array}} \right| = 5\)
\({A_{21}} = – \left| {\begin{array}{*{20}{c}} 2&5\\ 1&1 \end{array}} \right| = 3,\,{A_{22}} = \left| {\begin{array}{*{20}{c}} 1&5\\ { – 1}&1 \end{array}} \right| = 6\) and \({A_{23}} = – \left| {\begin{array}{*{20}{c}} 1&2\\ { – 1}&1 \end{array}} \right| = – 3\)
\({A_{31}} = – \left| {\begin{array}{*{20}{c}} 2&5\\ 3&1 \end{array}} \right| = – 13,\,{A_{32}} = – \left| {\begin{array}{*{20}{c}} 1&5\\ 2&1 \end{array}} \right| = 9\) and \({A_{33}} = \left| {\begin{array}{*{20}{c}} 1&2\\ 2&3 \end{array}} \right| = – 1\)
\(\therefore adj\,A = {\left[ {\begin{array}{*{20}{c}} 2&{ – 3}&5\\ 3&6&{ – 3}\\ { – 13}&9&{ – 1} \end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}} 2&3&{ – 13}\\ { – 3}&6&9\\ 5&{ – 3}&{ – 1} \end{array}} \right]\)
Now, \(\left({adj\,A} \right)A = \left[{\begin{array}{{c}} {21} & 0 & 0 \\ 0 & {21} & 0 \\ 0 & 0 & {21} \\ \end{array} } \right]\)
And \(A\left ({adj\,A} \right) = \left[{\begin{array}{{c}} {21} & 0 & 0 \\ 0 & {21} & 0 \\ 0 & 0 & {21} \\ \end{array} } \right]\)
Thus, \((adj\,A)A=|A|I=A(adj\,A)\)

Q.4: If \(A = \left[{\begin{array}{{c}} { – 4} & { – 3} & { – 3} \\ 1 & 0 & 1 \\ 4 & 4 & 3 \\ \end{array} } \right]\), Show that \(adj\,A = A\)
Ans:
Given that \(A = \left[{\begin{array}{{c}} { – 4} & { – 3} & { – 3} \\ 1 & 0 & 1 \\ 4 & 4 & 3 \\ \end{array} } \right]\)
Now,
\[{A_{11}} = \left| {\begin{array}{*{20}{c}} 0&1\\ 4&3 \end{array}} \right| = – 4,\,{A_{12}} = – \left| {\begin{array}{*{20}{c}} { 1}&{ 1}\\ 4&3 \end{array}} \right| = 1\] and \({A_{13}} = \left| {\begin{array}{*{20}{c}} 1&0\\ 4&4 \end{array}} \right| = 4\)
\({A_{21}} = \left| {\begin{array}{*{20}{c}} { – 3}&{ – 3}\\ 4&3 \end{array}} \right| = – 3,\,{A_{22}} = \left| {\begin{array}{*{20}{c}} { – 4}&{ – 3}\\ 4&3 \end{array}} \right| = 0\) and \({A_{23}} = – \left| {\begin{array}{*{20}{c}} { – 4}&{ – 3}\\ 4&4 \end{array}} \right| = 4\)
\({A_{31}} = \left| {\begin{array}{*{20}{c}} { – 3}&{ – 3}\\ 0&1 \end{array}} \right| = – 3,\,{A_{32}} = – \left| {\begin{array}{*{20}{c}} { – 4}&{ – 3}\\ 1&1 \end{array}} \right| = 1\) and \({A_{33}} = \left| {\begin{array}{*{20}{c}} { – 4}&{ – 3}\\ 1&0 \end{array}} \right| = 3\)
\(\therefore \,adj\,A = {\left[{\begin{array}{{c}} { – 4} & 1 & 4 \\ { – 3} & 0 & 4 \\ { – 3} & 1 & 3 \\ \end{array} } \right]^T} = \left[{\begin{array}{{c}} { – 4} & { – 3} & { – 3} \\ 1 & 0 & 1 \\ 4 & 4 & 3 \\ \end{array} } \right] = A\)

Q.5. If \(A = \left[{\begin{array}{{c}} { – 1} & { – 2} & { – 2} \\ 2 & 1 & { – 2} \\ 2 & { – 2} & 1 \\ \end{array} } \right]\), show that \(adj\,A=3A^T\)
Ans:
Given matrix is \(A = \left[{\begin{array}{{c}} { – 1} & { – 2} & { – 2} \\ 2 & 1 & { – 2} \\ 2 & { – 2} & 1 \\ \end{array} } \right]\)
Now, \({A_{11}} = – \left| {\begin{array}{*{20}{c}} 1&{ – 2}\\ { – 2}&1 \end{array}} \right| = – 3,\,{A_{12}} = – \left| {\begin{array}{*{20}{c}} 2&{ – 2}\\ 2&1 \end{array}} \right| = – 6\) and \({A_{13}} = \left| {\begin{array}{*{20}{c}} 2&1\\ 2&{ – 2} \end{array}} \right| = – 6\)
\({A_{21}} = – \left| {\begin{array}{*{20}{c}} { – 2}&{ – 2}\\ { – 2}&1 \end{array}} \right| = 6,\,{A_{22}} = \left| {\begin{array}{*{20}{c}} { – 1}&{ – 2}\\ 2&1 \end{array}} \right| = 3\) and \({A_{23}} = – \left| {\begin{array}{*{20}{c}} { – 1}&{ – 2}\\ 2&{ – 2} \end{array}} \right| = – 6\)
\({A_{31}} = \left| {\begin{array}{*{20}{c}} { – 2}&{ – 2}\\ 1&{ – 2} \end{array}} \right| = 6,\,{A_{32}} = – \left| {\begin{array}{*{20}{c}} { – 1}&{ – 2}\\ 2&{ – 2} \end{array}} \right| = – 6\) and \({A_{33}} = \left| {\begin{array}{*{20}{c}} { – 1}&{ – 2}\\ 2&1 \end{array}} \right| = 3\)
\(adj\,A = {\left[{\begin{array}{{c}} { – 3} & { – 6} & { – 6} \\ 6 & 3 & { – 6} \\ 6 & { – 6} & 3 \\ \end{array} } \right]^T} = \left[{\begin{array}{{c}} { – 3} & 6 & 6 \\ { – 6} & 3 & { – 6} \\ { – 6} & { – 6} & 3 \\ \end{array} } \right]\)
\({A^T} = \left[{\begin{array}{{c}} { – 1} & 2 & 2 \\ { – 2} & 1 & { – 2} \\ { – 2} & { – 2} & 1 \\ \end{array} } \right] \Rightarrow 3{A^T} = \left[{\begin{array}{{c}} { – 3} & 6 & 6 \\ { – 6} & 3 & { – 6} \\ { – 6} & { – 6} & 3 \\ \end{array} } \right]\)
\( \Rightarrow 3{A^T} = adj\,A\)

Summary

The adjoint of a matrix is one of the most powerful tools in mathematics. We have learnt about the adjoint of a matrix, its properties, and examples. The majority of tough problems can be simply handled with matrices. Solving linear equations and other mathematical functions such as calculus, optics, and quantum physics are all done with these instruments.

It has a wide range of applications in the real world, which has led to it playing a pivotal role in mathematics.

FAQs on Adjoint of a Matrix 2×2 and 3X3

Q.1: How do you find the adjoint of a matrix?
Ans: To find the adjoint of a matrix, we must first determine the cofactor of each element, followed by two more stages. The steps are listed below.
Step 1: Determine the cofactor for each element in the matrices.
Step 2: Using the cofactors, create a new matrix and expand the cofactors, resulting in a matrix.
Step 3: Now find the matrix’s transpose, which you got from Step 2.

Q.2. What is the adjoint of a \(2×2\) matrix?
Ans: For a square matrix of order \(2\), given by
\(A = \left[{\begin{array}{{c}} {{a_{11}}} & {{a_{12}}} \\ {{a_{21}}} & {{a_{22}}} \\ \end{array} } \right]\)
The \(adj\,A\) can also be obtained by interchanging \(a_{11}\) and \(a_{22}\) and by changing signs of \(a_{12}\) and \(a_{21}\), i.e.,
\(adj A = \left[{\begin{array}{{c}} {{a_{22}}} & {-{a_{12}}} \\ {-{a_{21}}} & {{a_{11}}} \\ \end{array} } \right]\)

Q.2: How do you find the adjoint of a \(3×3\) matrix?
Ans: \(A = \left[{\begin{array}{{c}} {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\ \end{array} } \right]\)
then \(adj(A) =C^T\), where \(C\) is the cofactor matrix of \(A\).

Q.4. What is the cofactor of a matrix?
Ans: In mathematics, a cofactor is used to find the adjoined inverse of a matrix. When you delete the column and row of a given element in a matrix, which is simply a numerical grid in the shape of a rectangle or a square, you get the Cofactor.
Let \(A = \left[{\begin{array}{{c}} {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\ \end{array} } \right]\)
Then \(adj\,A =\) Transpose of \(\left[{\begin{array}{{c}} {{A_{11}}} & {{A_{12}}} & {{A_{13}}} \\ {{A_{21}}} & {{A_{22}}} & {{A_{23}}} \\ {{A_{31}}} & {{A_{32}}} & {{A_{33}}} \\ \end{array} } \right]\)\(=\)\(\left[{\begin{array}{{c}} {{A_{11}}} & {{A_{21}}} & {{A_{31}}} \\ {{A_{12}}} & {{A_{22}}} & {{A_{32}}} \\ {{A_{13}}} & {{A_{23}}} & {{A_{33}}} \\ \end{array} } \right]\) 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 \(a_{ij}\).

Q.3: Is adjoint and transpose of a matrix the same?
Ans: No, it is not. Transposing a matrix simply means flipping the rows and columns, while the transposition of the matrix of cofactors is the adjoint of a matrix.

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