Transpose of a Matrix: Definition, Properties and Examples

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Gaurav Goplani

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The transpose of a matrix in linear algabra is one of the most widely used methods in matrix transformation. A matrix is a collection of a rectangular array of numbers or functions that are arranged in a specific number of rows and columns. The numbers or functions that are stored in a matrix are termed as the elements or the entries of the matrix. Matrix is denoted by capital letters enclosed within square brackets, and can be represented as vertical matrix, horizontal matrix and square matrix. The transpose of a matrix for a given matrix is obtained by interchanging rows into columns or columns to rows. Let’s learn more on transpose of a Matrix and discuss some important questions. 

Keyterms: Matrix, Matrix transformation, Vertical matrix, Horizontal matrix, Square matrix, Linear algabra, Columns, Rows

Read more: Statistics Formula


What is a Matrix?

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The numbers or functions that are stored in a matrix are termed as the elements or the entries of the matrix. Matrix is denoted by capital letters enclosed within square brackets, and can be represented as follows:

Elements in the Matrix
Elements in the Matrix

Here, three horizontal elements in the Matrix that are called the Columns, and two vertical elements that are called the Rows. Hence, in this example, the matrix comprises 2 rows and 3 columns. Therefore, the matrix is denoted as a 2x3 matrix.

  • If the number of rows is greater than the number of columns, then it is called a vertical matrix
  • If the number of columns is greater than the number of rows, then it is called a horizontal matrix
  • If the number of rows and columns are equal, then it is called a square matrix.

Read Also: Area of a Triangle


Transpose Matrix

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The matrix obtained by interchanging the rows and columns of the parent matrix is called the Transpose matrix. The definition of a transpose matrix goes like this - “A Matrix which is formed by turning all the rows of a given matrix into columns and vice-versa.”

Transpose matrix of A is denoted by A’. It can be better understood by the following example:

\(A = \begin{bmatrix}1 & 5 & 6 \\[0.3em]8 & 10 & 11 \\[0.3em]22 & 9 & 4 \\[0.3em]5 & 7 & 9 \\[0.3em] \end{bmatrix} \)
\(A' = \begin{bmatrix}1 & 8 & 10 & 11 \\[0.3em]5 & 10 & 9 & 7 \\[0.3em]6 & 11 & 4 & 9 \\[0.3em]\end{bmatrix} \)

Transpose matrix of A is denoted by A’

Here, in Matrix A, the number of rows was 4 and the number of columns was 3 but, on taking the transpose of A we obtained A’ having 3 rows and 4 columns. Thus, the vertical Matrix gets converted into Horizontal Matrix.

Hence, we can say if the matrix before taking the transpose was a vertical matrix, it will be transposed to a horizontal matrix and vice-versa.

Read More: Minors and Cofactors


Properties of Transpose of the Matrices

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To the understanding clear, we shall take an example in each case as under:

Transpose of the Transpose Matrix 

The transpose of a transpose of a Matrix is the original Matrix itself. It can be clearly understood that interchanging the rows and columns twice will result in the original Matrix itself. Mathematically, it can be written as,

(A’)’ = A

Let us take an example to understand this property:

Consider a Matrix A as follows: A = \(\begin{bmatrix}3 & \sqrt{3} & 2 \\[0.3em]4 & 2 & 0 \\[0.3em] \end{bmatrix}\)⇒ A’ = \(\begin{bmatrix}3 & 4 \\[0.3em] \sqrt{3} & 2\\[0.3em] 2 & 0 \\[0.3em]\end{bmatrix}\)⇒ (A’)’ = \(\begin{bmatrix}3 & \sqrt{3} & 2 \\[0.3em]4 & 2 & 0 \\[0.3em] \end{bmatrix}\) = A

Therefore, it can be written as: (A’)’ = A.

Addition Properties of Matrix 

The addition of the transpose of Matrices is equal to the transpose of the result of the addition of the original Matrices. Mathematically, it can be written as,

(A + B)’ = A’ + B’

Let us consider an example to understand this property in more detail:

Consider 2 matrices A & B as follows: A = \(\begin{bmatrix}3 & \sqrt{3} & 2 \\[0.3em]4 & 2 & 0 \\[0.3em] \end{bmatrix}\), B = \(\begin{bmatrix}2 & -1 & 2 \\[0.3em]1 & 2 & 4 \\[0.3em] \end{bmatrix} \) ⇒ A + B = \(\begin{bmatrix}5 & \sqrt{3} -1 & 4 \\[0.3em]5 & 4 & 4 \\[0.3em] \end{bmatrix} \)

Now, the transpose of their addition (A+B)’ can be written as follows: A =\(\begin{bmatrix}3 & \sqrt{3} & 2 \\[0.3em]4 & 2 & 0 \\[0.3em] \end{bmatrix}\), B =\(\begin{bmatrix}2 & -1 & 2 \\[0.3em]1 & 2 & 4 \\[0.3em] \end{bmatrix} \)⇒ A + B = \(\begin{bmatrix}5 & \sqrt{3} -1 & 4 \\[0.3em]5 & 4 & 4 \\[0.3em] \end{bmatrix} \)

Now, let us add the transpose of the Matrices A & B together to get A’ + B’,

A = \(\begin{bmatrix}3 & \sqrt{3} & 2 \\[0.3em]4 & 2 & 0 \\[0.3em] \end{bmatrix}\)B =\(\begin{bmatrix}2 & -1 & 2 \\[0.3em]1 & 2 & 4 \\[0.3em] \end{bmatrix} \)⇒ A + B = \(\begin{bmatrix}5 & \sqrt{3} -1 & 4 \\[0.3em]5 & 4 & 4 \\[0.3em] \end{bmatrix} \)

\((A + B)' = \begin{bmatrix}5 & 5 \\[0.3em] \sqrt{3} - 1 & 4\\[0.3em] 4 & 4 \\[0.3em]\end{bmatrix}\)

From the above example, we can see clearly that (A+B)’ = A’ + B’.

Multiplication by Constant

The transpose of the multiplication of a Matrix with a constant is always equal to the multiplication of that constant with the transpose of the original Matrix. Mathematically, this property can be written as follows:

kA’ = (kA)’

Let us take an example to understand this property:

Consider a Matrix B and constant k as follows:

\(kB = k \begin{bmatrix}2 & -1 & 2 \\[0.3em]1 & 2 & 4 \\[0.3em] \end{bmatrix} = \begin{bmatrix}2k & -k & 2k \\[0.3em]k & 2k & 4k \\[0.3em] \end{bmatrix} \)

Thus, it is proved that kB’ = (kB)’.

\((kB)' = \begin{bmatrix}2k & k \\[0.3em]-k & 2k \\[0.3em]2k & 4k \\[0.3em] \end{bmatrix} = k \begin{bmatrix}2 & 1 \\[0.3em]-1 & 2 \\[0.3em]2 & 4 \\[0.3em] \end{bmatrix}= kB' \)

Multiplication Property of Transpose

The transpose of the multiplication of two matrices is equal to the multiplication of transposes of the individual matrices in the reverse order. The Multiplication Property of transpose of matrices can be written as follows:

(AB)’ = B’A’

This can be easily understood by taking examples of two Matrices - A & B as follows

\(A = \begin{bmatrix}- 2 \\[0.3em]4 \\[0.3em]5 \\[0.3em] \end{bmatrix}, B =\begin{bmatrix}1 & 3 & -6 \\[0.3em]\end{bmatrix} \)

Now, multiplying the Matrices A & B with each other, we get

\(AB = \begin{bmatrix}- 2 \\[0.3em]4 \\[0.3em]5 \\[0.3em] \end{bmatrix}\begin{bmatrix}1 & 3 & -6 \\[0.3em]\end{bmatrix} =\begin{bmatrix}-2 & -6 & 12 \\[0.3em]4 & 12 & -24 \\[0.3em] 5 & 15 & -30 \\[0.3em]\end{bmatrix} \)

Now, transposing and multiplying each of the matrices,

\(A' =\begin{bmatrix}-2 & 4 & 5 \\[0.3em]\end{bmatrix} , B' = \begin{bmatrix}1 \\[0.3em]3 \\[0.3em]-6 \\[0.3em] \end{bmatrix}\)

\(B'A' = \begin{bmatrix}1 \\[0.3em]3 \\[0.3em]-6 \\[0.3em] \end{bmatrix}\begin{bmatrix}-2 & 4 & 5 \\[0.3em]\end{bmatrix}= \begin{bmatrix}-2 & 4 & 5 \\[0.3em]-6 & 12 & 15 \\[0.3em] 12 & -24 & -30 \\[0.3em]\end{bmatrix} = (AB)' \)

Thus, it can be written as (AB)’ = B’A’.

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Things to Remember

  • Rows and columns are interchanged during the transposition of matrices.
  • The Transpose Matrix of a Horizontal Matrix is Vertical and vice versa and that of the square Matrix remains square.
  • The following set of equations should also be remembered about the transposition of Matrix:
  • (A’)’ = A
  • (A + B)’ = A’ + B’
  • kA’ = (kA)’
  • (AB)’ = B’A’

Sample Questions

Ques. The transpose of a horizontal matrix is a ________. (2 marks)
(A) Horizontal matrix
(B) Vertical Matrix
(C) Square Matrix
(D) Can not be determined

Ans. B

Explanation: The transpose of a horizontal matrix is a vertical matrix and vice versa. One must also remember that the transpose of a square matrix is always square as the number of rows and columns remains equal.

Ques. If A’B’ = C’, then C = ________. (3 marks)
(A) AC
(B) BA
(C) BC
(D) ABC

Ans. B

Explanation: From the Multiplication Properties of Transpose of Matrix,

We know that,

(AB)’ = A’B’

From the given question,

A’B’=C’

(AB)’=C (Replacing A’B’ with (AB)’)

Now, taking transpose on both the sides,

((BA)’)’ = C’

BA = C’

Ques. Which of the following is not true generally according to the properties of transposes of matrices? (2 marks)
(A) A’B’ = (AB)’
(B) (A+B)’ = A’ + B’
(C) (A’)’ = A’
(D) kA’ = (kA)’

Ans. C

Explanation: According to the properties of transposes of Matrices, the transpose of the transpose of a Matrix is always equal to the original Matric, therefore Option C is incorrect. The correct equation would be : (A’)’ =

Ques. How to calculate the transpose of a matrix? (3 marks)

Ans. In order to calculate the transpose of a matrix, we must simply interchange the rows and columns of the matrix, that is write the elements of the rows as columns and write the elements of a column as rows. For instance, if a matrix is a 2×3 matrix, this means it has 2 rows and 3 columns. The elements in the first row of the given matrix are written in the first column of the new matrix while finding the transpose of a matrix. In the same way, the elements in the second row of the given matrix are written in the second column of the new matrix. Therefore, the order of the new matrix becomes 3×2, as it has 3 rows and 2 columns.

Ques. What is the additional property of transpose? (3 mrks)

Ans. The addition of the transpose of Matrices is equal to the transpose of the result of the addition of the original Matrices. Mathematically, it can be written as,

(A + B)’ = A’ + B’

Let us consider an example to understand this property in more detail:

Consider 2 matrices A & B as follows: A

A=\(\begin{bmatrix}3 & \sqrt{3} & 2 \\[0.3em]1 & 2 & 0 \\[0.3em] \end{bmatrix}, B =\begin{bmatrix}2 & -1 & 2 \\[0.3em]1 & 2 & 4 \\[0.3em] \end{bmatrix} \)  ⇒ A + B = \(\begin{bmatrix}5 & \sqrt{3} -1 & 4 \\[0.3em]5 & 4 & 4 \\[0.3em] \end{bmatrix} \)

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    Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).


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            Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).


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                    The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]


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                        Find:
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                          • \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
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                          • \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
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