 # Transpose of a Matrix

In Linear algebra, the transpose of a matrix is one of the most commonly used methods in matrix transformation. For a given matrix, the transpose of a matrix is obtained by interchanging rows into columns or columns to rows. In this article, we are going to learn the definition of the transpose of a matrix, steps to find the transpose of a matrix, properties and examples with a complete explanation.

Before learning how to find the transpose of a matrix, first let us learn, what a matrix is?

## 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.

$$\begin{array}{l}A = \begin{bmatrix} 2 & 13\\ -9 & 11\\ 3 & 17 \end{bmatrix}_{3 \times 2}\end{array}$$

The above matrix A is of order 3 × 2. Thus, there are a total of 6 elements.

The horizontal array is known as rows and the vertical array is known as Columns.

Now, let us take another matrix.

$$\begin{array}{l}B = \begin{bmatrix} 2 & -9 & 3\\ 13 & 11 & 17 \end{bmatrix}_{2 \times 3}\end{array}$$

The number of rows in matrix A is greater than the number of columns, such a matrix is called a Vertical matrix.

The number of columns in matrix B is greater than the number of rows. Such a matrix is called a Horizontal matrix.

One thing to notice here, if elements of A and B are listed, they are the same in number and each element that 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 the same and each corresponding element of P is equal to that of Q.

That is, if

$$\begin{array}{l}P\end{array}$$
=
$$\begin{array}{l} [p_{ij}]_{m×n}\end{array}$$
and
$$\begin{array}{l}Q\end{array}$$
=
$$\begin{array}{l} [q_{ij}]_{r×s}\end{array}$$
are two matrices such that
$$\begin{array}{l} P\end{array}$$
=
$$\begin{array}{l}Q\end{array}$$
, then:

• m = r and n = s i.e. the orders of the two matrices must be same
• For every value of i and j,
$$\begin{array}{l}p_{ij}\end{array}$$
=
$$\begin{array}{l}q_{ij}\end{array}$$
.

## Transpose of a Matrix Definition

The transpose of a matrix is found by interchanging its rows into columns or columns into rows. The transpose of the matrix is denoted by using the letter “T” in the superscript of the given matrix. For example, if “A” is the given matrix, then the transpose of the matrix is represented by A’ or AT.

The following statement generalizes the matrix transpose:

If

$$\begin{array}{l}A\end{array}$$
=
$$\begin{array}{l}[a_{ij}]_{m×n}\end{array}$$
, then
$$\begin{array}{l}A'\end{array}$$
=
$$\begin{array}{l}[a_{ij}]_{n×m}\end{array}$$
.

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 vice-versa.”

## How to Find the Transpose of a Matrix?

Consider an example, if a matrix is a 2×3 matrix. It means it has 2 rows and 3 columns. While finding the transpose of a matrix, the elements in the first row of the given matrix are written in the first column of the new matrix. Similarly, the elements in the second row of the given matrix are written in the second column of the new matrix. Hence, the order of the new matrix becomes 3×2, as it has 3 rows and 2 columns. Let’s Work Out- Example- Find the transpose of the given matrix $$\begin{array}{l}M = \begin{bmatrix} 2 & -9 & 3 \\ 13 & 11 & -17 \\ 3 & 6 & 15 \\ 4 & 13 & 1 \end{bmatrix} \end{array}$$ Solution- Given a matrix of the order 4×3. The transpose of a matrix is given by interchanging rows and columns. $$\begin{array}{l}M^T = \begin{bmatrix} 2 & 13 & 3 & 4 \\ -9 & 11 & 6 & 13\\ 3 & -17 & 15 & 1 \end{bmatrix}\end{array}$$

## Properties of Transpose of a Matrix

To understand the properties of the matrix transpose, we will take two matrices A and B which have equal order. Some 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,

$$\begin{array}{l}(A’ )'\end{array}$$
=
$$\begin{array}{l}A\end{array}$$

What basically happens, is that any element of A, i.e.

$$\begin{array}{l}a_{ij}\end{array}$$
gets converted to
$$\begin{array}{l}a_{ji}\end{array}$$
if the transpose of A is taken. So, taking transpose again, it gets converted to
$$\begin{array}{l}a_{ij}\end{array}$$
, which was the original matrix
$$\begin{array}{l}A\end{array}$$
.

 Example: If $$\begin{array}{l}N = \begin{bmatrix} 22 & -21 & -99 \\ 85 & 31 & -2\sqrt{3} \\ 7 & -12 & 57 \end{bmatrix}\end{array}$$, Then $$\begin{array}{l}N’ = \begin{bmatrix} 22 &85 & 7 \\ -21 & 31 & -12 \\ -99 & -2\sqrt{3} & 57 \end{bmatrix}\end{array}$$ Now, $$\begin{array}{l}(N’)'\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 22 & -21 & -99 \\ 85 & 31 & -2\sqrt{3} \\ 7 & -12 & 57 \end{bmatrix} \end{array}$$ = $$\begin{array}{l}N\end{array}$$

### (ii) Addition Property of Transpose

Transpose of an addition of two matrices A and B obtained will be exactly equal to the sum of the transpose of individual matrices A and B.

This means,

$$\begin{array}{l}(A+B)'\end{array}$$
=
$$\begin{array}{l}A’+B'\end{array}$$

 Example-  If$$\begin{array}{l} P\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 2 & -3 & 8 \\ 21 & 6 & -6 \\ 4 & -33 & 19 \end{bmatrix} \end{array}$$ and $$\begin{array}{l}Q\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 1 & -29 & -8 \\ 2 & 0 & 3 \\ 17 & 15 & 4 \end{bmatrix} \end{array}$$ $$\begin{array}{l}P + Q\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 2+1 & -3-29 & 8-8 \\ 21+2 & 6+0 & -6+3 \\ 4+17 & -33+15 & 19+4 \end{bmatrix} \end{array}$$= $$\begin{array}{l} \begin{bmatrix} 3 & -32 & 0 \\ 23 & 6 & -3 \\ 21 & -18 & 23 \end{bmatrix} \end{array}$$ $$\begin{array}{l}(P+Q)'\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & -18 \\ 0 & -3 & 23 \end{bmatrix} \end{array}$$ $$\begin{array}{l}P’+Q'\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 2 & 21 & 4 \\ -3 & 6 & -33 \\ 8 & -6 & 19 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 17 \\ -29 & 0 & 15 \\ -8 & 3 & 4 \end{bmatrix} \end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & -18 \\ 0 & -3 & 23 \end{bmatrix} \end{array}$$ = $$\begin{array}{l}(P+Q)'\end{array}$$ So, we can observe that $$\begin{array}{l}(P+Q)'\end{array}$$ = $$\begin{array}{l}P’+Q'\end{array}$$.

### (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 the original matrix multiplied by that constant. That is,

$$\begin{array}{l}(kA)'\end{array}$$
=
$$\begin{array}{l}kA'\end{array}$$
, where k is a constant

 Example- If $$\begin{array}{l}P\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 2 & 8 & 9 \\ 11 & -15 & -13 \end{bmatrix}_{2×3} \end{array}$$ and k is a constant, then $$\begin{array}{l}(kP)'\end{array}$$= $$\begin{array}{l} \begin{bmatrix} 2k & 11k \\ 8k & -15k \\ 9k &-13k \end{bmatrix}_{2×3} \end{array}$$ $$\begin{array}{l}kP'\end{array}$$= $$\begin{array}{l} k \begin{bmatrix} 2 & 11 \\ 8 & -15 \\ 9 & -13 \end{bmatrix}_{2×3} \end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 2k & 11k \\ 8k & -15k \\ 9k &-13k \end{bmatrix}_{2×3} \end{array}$$ = $$\begin{array}{l}(kP)'\end{array}$$ We can observe that $$\begin{array}{l}(kP)'\end{array}$$ = $$\begin{array}{l}kP'\end{array}$$.

### (iv) Multiplication Property of Transpose

Transpose of the product of two matrices is equal to the product of transpose of the two matrices in reverse order. That is

$$\begin{array}{l}(AB)'\end{array}$$
=
$$\begin{array}{l}B’A'\end{array}$$

 Example: $$\begin{array}{l}A \end{array}$$= $$\begin{array}{l} \begin{bmatrix} 9 & 8 \\ 2 & -3 \end{bmatrix} \end{array}$$ and $$\begin{array}{l} B \end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 4 & 2 \\ 1 & 0 \end{bmatrix} \end{array}$$ Let us find $$\begin{array}{l}A×B\end{array}$$. $$\begin{array}{l}A×B\end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 44 & 18 \\ 5 & 4 \end{bmatrix} \Rightarrow (AB)’ = \begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix} \end{array}$$ $$\begin{array}{l}B’A'\end{array}$$ = $$\begin{array}{l}\begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix} \end{array}$$ = $$\begin{array}{l} \begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix} \end{array}$$ = $$\begin{array}{l}(AB)'\end{array}$$ ∴ $$\begin{array}{l}(AB)'\end{array}$$ = $$\begin{array}{l}B’A'\end{array}$$ $$\begin{array}{l}A’B'\end{array}$$ = $$\begin{array}{l}\begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 40 & 9 \\ 26 & 8 \end{bmatrix}\end{array}$$ 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.

## Transpose of a Matrix Video Lesson

### Transpose of a Matrix ### Transpose of a Matrix Examples

Go through the following problems to understand how to find the transpose of a matrix.

Example 1:

If matrix

$$\begin{array}{l}A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}\end{array}$$
. Find the transpose of matrix A.

Solution:

Given:

Matrix

$$\begin{array}{l}A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}\end{array}$$

On interchanging the rows and columns of the given matrix, the transpose of matrix A is given as:

$$\begin{array}{l}A^{T} = \begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix}\end{array}$$
.

Therefore, the transpose of matrix A,

$$\begin{array}{l}A^{T} = \begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix}\end{array}$$

Example 2:

Find the transpose for the given 2×2 matrix,

$$\begin{array}{l}X =\begin{bmatrix} 7 & 11 \\ 21 & 16 \end{bmatrix}\end{array}$$

Solution:

Given 2×2 matrix,

$$\begin{array}{l}X =\begin{bmatrix} 7 & 11 \\ 21 & 16 \end{bmatrix}\end{array}$$

Hence, the transpose of the given 2×2 matrix is:

$$\begin{array}{l}X^{T} =\begin{bmatrix} 7 & 21 \\ 11 & 16 \end{bmatrix}\end{array}$$

### Practice Problems

1. Give matrix
$$\begin{array}{l}A=\begin{bmatrix}1 & 2 & 2\\2 & 1 & -2\\4 & -3 & 1\end{bmatrix}\end{array}$$
. Find the transpose of A and then prove that (AT)T = A.
2. Verify the addition property for the transpose of matrices
$$\begin{array}{l}P=\begin{bmatrix}4 & 0 & 3\\5 & -1 & -2\\1 & 2 & 1\end{bmatrix}\textrm{and}\ Q=\begin{bmatrix}5 & 1 & 1\\3 & -4 & 0\\2 & 1 & -3\end{bmatrix}\end{array}$$
.
3. Find AT for the matrix
$$\begin{array}{l}A=\begin{bmatrix}1 & -4 & 9\\5 & 3 & -2\end{bmatrix}\end{array}$$
.

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### What is the transpose of a matrix?

The transpose of a matrix can be defined as an operator which can switch the rows and column indices of a matrix i.e. it flips a matrix over its diagonal.

### How to calculate the transpose of a Matrix?

To calculate the transpose of a matrix, simply interchange the rows and columns of the matrix i.e. write the elements of the rows as columns and write the elements of a column as rows.

### What is the Addition Property of Transpose?

The addition property of transpose is that the sum of two transpose matrices will be equal to the sum of the transpose of individual matrices. So,

• (A+B)′ = A′+B′

### What is the Multiplication Property of Transpose?

The multiplication property of transpose is that the transpose of a product of two matrices will be equal to the product of the transpose of individual matrices in reverse order. So,

• (A×B)′ = B′ × A′