What is a matrix?
A matrix is a rectangular array of numbers or functions arranged in a fixed number of rows and columns.
There are many types of matrices. Let us consider a matrix to understand more about them.
\(A = \begin{bmatrix} 2 & 13\\ 9 & 11\\ 3 & 17 \end{bmatrix}_{3 \times 2}\)
The above matrix A is of order 3 × 2. Thus there are total 6 elements.
The horizontal array are known as Rows and the vertical array are known as Columns.
Now, let us take another matrix.
\(B = \begin{bmatrix} 2 & 9 & 3\\ 13 & 11 & 17 \end{bmatrix}_{2 \times 3}\)
The number of rows in matrix A is greater than the number of columns, such a matrix is called as a Vertical matrix.
The number of columns in matrix B is greater than the number of rows. Such a matrix is called as a Horizontal matrix.
One thing to notice here, if elements of A and B are listed, they are same in number and each element which is there in A is there in B too. So, is A = B?
Before answering this, we should know how to decide the equality of the matrices.
A matrix P is said to be equal to matrix Q if their orders are same and each corresponding element of P is equal to that of Q.
That is, if \(P\) =\( [p_{ij}]_{m×n}\) and \(Q\) =\( [q_{ij}]_{r×s}\) are two matrices such that\( P\) = \(Q\), then:
 m = r and n = s i.e. the orders of the two matrices must be same
 For every value of i and j, \(p_{ij}\) = \(q_{ij}\).
Transpose of a matrix:
Let us now go back to our original matrices A and B. Though they have the same set of elements, are they equal?
The answer is no. That’s because their order is not same. Now, there is an important observation. There can be many matrices which have exactly the same elements as A has.
Here, the number of rows and columns in A is equal to number of columns and rows in B respectively. Thus the matrix B is known as the Transpose of the matrix A. The transpose of matrix A is represented by \(A’\) or \(A^T\). The following statement generalizes transpose of a matrix:
If \(A\) = \([a_{ij}]_{m×n}\), then \(A’\) =\([a_{ij}]_{n×m}\).
Thus Transpose of a Matrix is defined as “A Matrix which is formed by turning all the rows of a given matrix into columns and viceversa.”
Let’s Work Out Example Find the Transpose of the given matrix \(M = \begin{bmatrix} 2 & 9 & 3 \\ 13 & 11 & 17 \\ 3 & 6 & 15 \\ 4 & 13 & 1 \end{bmatrix} \) Solution Given matrix of the order 4×3. The transpose of a matrix is given by interchanging of rows and columns. \(M^T = \begin{bmatrix} 2 & 13 & 3 & 4 \\ 9 & 11 & 6 & 13\\ 3 & 17 & 15 & 1 \end{bmatrix}\) 
Properties of Transpose of a Matrix:
To understand the properties of a transpose matrix, we will take two matrices A and B which have equal order. Some of the properties of the transpose of a matrix are given below:
(i) Transpose of the Transpose Matrix:
If we take the transpose of the transpose matrix, the matrix obtained is equal to the original matrix. Hence, for a matrix A,
\((A’ )’\) = \(A\)
What basically happens, is that any element of A, i.e. \(a_{ij}\) gets converted to \(a_{ji}\) if transpose of A is taken. So, taking the transpose again, it gets converted to \(a_{ij}\), which was the original matrix \(A\).
Example: If \(N = \begin{bmatrix} 22 & 21 & 99 \\ 85 & 31 & 2\sqrt{3} \\ 7 & 12 & 57 \end{bmatrix}\), Then \(N’ = \begin{bmatrix} 22 &85 & 7 \\ 21 & 31 & 12 \\ 99 & 2\sqrt{3} & 57 \end{bmatrix}\) Now, \((N’)’\) = \( \begin{bmatrix} 22 & 21 & 99 \\ 85 & 31 & 2\sqrt{3} \\ 7 & 12 & 57 \end{bmatrix} \) = \(N\) 
(ii) Addition Property of Transpose:
The Transpose of an addition of two matrices A and B obtained will be exactly equal to the sum of the transpose of individual matrix A and B.
This means,
\((A+B)’\) = \(A’+B’\)
Example If\( P\) = \( \begin{bmatrix} 2 & 3 & 8 \\ 21 & 6 & 6 \\ 4 & 33 & 19 \end{bmatrix} \)

(iii) Multiplication by Constant:
If a matrix is multiplied by a constant and its transpose is taken, then the matrix obtained is equal to the transpose of original matrix multiplied by that constant. That is,
\((kA)’\) = \(kA’\), where k is a constant
Example If \(P\) = \( \begin{bmatrix} 2 & 8 & 9 \\ 11 & 15 & 13 \end{bmatrix}_{2×3} \) and k is a constant, then \((kP)’\)= \( \begin{bmatrix} 2k & 11k \\ 8k & 15k \\ 9k &13k \end{bmatrix}_{2×3} \)\(kP’\)= \( k \begin{bmatrix} 2 & 11 \\ 8 & 15 \\ 9 & 13 \end{bmatrix}_{2×3} \) = \( \begin{bmatrix} 2k & 11k \\ 8k & 15k \\ 9k &13k \end{bmatrix}_{2×3} \) = \((kP)’\) We can observe that \((kP)’\) = \(kP’\). 
(iv) Multiplication Property of Transpose:
The Transpose of the product of two matrices is equal to the product of the transpose of the two matrices in reverse order. That is
\((AB)’\) = \(B’A’\)
Example: \(A \)= \( \begin{bmatrix} 9 & 8 \\ 2 & 3 \end{bmatrix} \) and \( B \) = \( \begin{bmatrix} 4 & 2 \\ 1 & 0 \end{bmatrix} \) Let us find \(A×B\). \(A×B\) = \( \begin{bmatrix} 44 & 18 \\ 5 & 4 \end{bmatrix} \Rightarrow (AB)’ = \begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix} \) \(B’A’\) = \(\begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 9 & 2 \\ 8 & 3 \end{bmatrix} \) = \( \begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix} \) = \((AB)’\) ∴ \((AB)’\) = \(B’A’\)
\(A’B’\) = \(\begin{bmatrix} 9 & 2 \\ 8 & 3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 40 & 9 \\ 26 & 8 \end{bmatrix}\) We can clearly observe from here that (AB)’≠A’B’. 
Those were properties of matrix transpose which are used to prove several theorems related to matrices.
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