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Transpose of a Matrix

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Transpose of matrix is a very common method used for matrix transformation in linear algebra. Transpose of matrix is obtained by interchanging the rows and columns of the given matrix or vice versa. Transpose of a matrix can be utilized to obtain the adjoint and inverse of the matrices. Before learning about the details of the transpose of a matrix let’s first learn about “What is a matrix?”. A matrix is nothing but the representation of the set of data in the rectangular array format. In a matrix, data is arranged in specific rows and columns. Various types of matrices exist in Mathematics and are presented in the order of rows × columns. Let’s take an example of the matrix of order 3 × 2 (say A).

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

What is the Transpose of a Matrix?

Transpose of a matrix is a matrix that is obtained by swapping the rows and columns of the given matrix or vice versa, i.e., for the given matrix the elements in rows are interchanged with the elements in columns. For any given matrix A its transpose is denoted as At, or AT.

Let, A is a matrix of order m × n then At be the transpose of matrix A with order n × m
where,

A = [a(ij)]m × n 
At = [a(ji)]n × m 

here i, j present the position of a matrix element, row- and column-wise, respectively, such that,1 ≤ i ≤ m and 1 ≤ j ≤ n.

Example: For any given matrix A of order 2 × 3 its transpose is?

 A = \begin{bmatrix} 2 & 5 & 3\\ 4 & 7 & 0 \end{bmatrix}

Solution:

Transpose of A

At\begin{bmatrix} 2 & 4 \\ 5 & 7 \\ 3 & 0 \end{bmatrix}

Order of At is 3 × 2

Symbol of Transpose Matrix

Transpose of a matrix is the operation that flips the matrix over its principal diagonal and interchanges its rows with columns. Transpose of a matrix A is denoted by the notation A’ or AT or At.

Order of Transpose Matrix

The order of a matrix tells the total elements that a matrix contains. It also represents the number of rows and columns in a matrix. Horizontal values represent the rows of the matrix, and vertical values represent the columns of the matrix. For any matrix Am×n, the order is m×n, i.e., it has m rows and n columns. Therefore, the transpose of matrix A is At and its order is n×m, i.e., it has n rows and m columns.

How to Find the Transpose of a Matrix?

Transpose of any matrix can easily be found by changing the values in the rows with the values in the columns. Let’s take an example to understand this in detail.

For any matrix A2×3, the order is 2×3 which means it has 2 rows and 3 columns.

A =   \begin{bmatrix} a & b & c\\ x & y & z \end{bmatrix}

The transpose of matrix A is At of the order 3×2 having 3 rows and 2 columns. In the transpose matrix elements of the first row of the given matrix are changed with the first column of the transpose matrix. Similarly, the elements of the second row of the given matrix A are swapped with the second column of the new matrix At and so on till the whole matrix is swapped.

At\begin{bmatrix} a & x \\ b & y \\ c & z \end{bmatrix}

Transpose of Row and Column Matrices

A matrix that has a single row is known as a row matrix, whereas a matrix that has a single column is known as a column matrix. The transpose of a row matrix is a column matrix and vice versa. For example, if P is a column matrix of order “4 × 1,” then its transpose is a row matrix of order “1 × 4.” If Q is a row matrix of order “1 × 3,” then its transpose is a column matrix of order “3 × 1.”

P = \left[\begin{array}{cccc} a & b & c & d\end{array}\right]⇒ P^{t} = \left[\begin{array}{c} a\\ b\\ c\\ d \end{array}\right]

Q = \left[\begin{array}{c} p\\ q\\ r \end{array}\right]⇒ Q^{t} = \left[\begin{array}{ccc} p & q & r\end{array}\right]

Transpose of Horizontal and Vertical Matrices

If the number of rows in a matrix is less than the number of columns, then the matrix is known as a horizontal matrix, and if the number of columns in a matrix is less than the number of rows, then the matrix is known as a vertical matrix. The transpose of a horizontal matrix is a vertical matrix and vice versa. For example, if M is a horizontal matrix of order “2 × 3,” then its transpose is a vertical matrix of order “3 × 2.”

M = \left[\begin{array}{ccc} 2 & 0 & -1\\ 0 & 3 & 4 \end{array}\right]_{2\times3}⇒ M^{t} = \left[\begin{array}{cc} 2 & 0\\ 0 & 3\\ -1 & 4 \end{array}\right]_{3\times2}

N = \left[\begin{array}{ccc} 2 & 3 & 4\\ 4 & 6 & 8\\ 6 & 9 & 12\\ 8 & 12 & 16 \end{array}\right]_{4\times3}⇒ N^{t} = \left[\begin{array}{cccc} 2 & 4 & 6 & 8\\ 3 & 6 & 9 & 12\\ 4 & 8 & 12 & 16 \end{array}\right]_{3\times4}

Transpose of a Square Matrix

Square matrices are matrices that have an equal number of rows and columns. for any square matrix An×n, its transpose has the same order i.e., the transpose of A, At has order n × n. The rows and columns are interchanged in the transpose of a square matrix.

Transpose of a 2 × 2 Matrix

For any 2 × 2 matrices A, 

A = \begin{bmatrix} a & x \\ b & y \end{bmatrix}

its transpose is At,

At = \begin{bmatrix} a & b \\ x & y \end{bmatrix}

Example: Find the transpose of matrix A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

Solution:

Transpose of the matrix A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}     is

At =  \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}

Transpose of a 3 × 3 Matrix

For any 3 × 3 matrices A, 

A = \begin{bmatrix} a & x & p \\ b & y & q \\ c & z & r \end{bmatrix}

its transpose is At,

At = \begin{bmatrix} a & b & c \\ x & y & z \\ p & q & r \end{bmatrix}

Example: Find the transpose of matrix A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}

Solution:

Transpose of the matrix A =\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}     is

At\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix}

Determinant of Transpose of a Matrix

The determinant of the transpose of a matrix A is equal to the determinant of A itself, i.e., for any square matrix A

|A| = |AT|

Properties of Transpose of a Matrix

Let’s learn about the important properties of the transpose of a matrix:

  • A square matrix “A” of order “n × n” is said to be an orthogonal matrix, if AAT = ATA = I, where “I” is an identity matrix of order “n × n.”
  • A square matrix “A” of order “n × n” is said to be a symmetric matrix if its transpose is the same as the original matrix, i.e., AT = A.
  • A square matrix “A” of order “n × n” is said to be a skew-symmetric matrix if its transpose is equal to the negative of the original matrix, i.e., AT = –A.
  • Double Transpose of a Matrix: Transpose of the transpose matrix is the original matrix itself.

(At)t = A

  • Transpose of Product of Matrices: This property says that 

(AB)t = BtAt

Proof:

If matrices A and B are of orders m × n and n × p, respectively.

and 

At and Bt are the transpose of matrices A and B of orders n × m and p × n respectively (from the product rule of matrices).

It implies, if A = [a(ij)], and At = [c(ji)]

Then, [c(ji)] = [a(ij)]

and,

If B = [b(jk)], and Bt = [d(kj)]

Then, [d(kj)] = [b(jk)]

Now, from the product rule of matrices, we can write,

AB is m × p matrix and (AB)t is p × m matrix.

Also, Bt is a p × n matrix, and At is an n × m matrix.

This implies that,

(Bt)(At) is a p × m matrix.

Therefore,

(AB)t and (Bt)(At) are both p × m matrices.

Now we can write,

(k, i)th element of (AB)t = (i, k)th element of AB 

\sum_{j=1}^{n}  a_{ij}  b_{jk}
\sum_{j=1}^{n}  c_{ji}  d_{kj}

\sum_{j=1}^{n}  d_{kj}  c_{ji}

(k, i)th element of (Bt)(At)

Therefore, 

the elements of (AB)t and (Bt)(At) are equal.

Therefore,

(AB)t = (Bt)(At)

  • Multiplication by Constant: If a matrix is multiplied by a scalar value and its transpose is taken, then the resultant matrix will be equal to the transpose of the original matrix multiplied by the scalar value, i.e., (kA)t = kAt, where k is a scalar value.

Proof:

Let us consider a matrix A = [aij]m × n and a scalar k.

The order of the given matrix A is m × n.

If matrix A is multiplied by the scalar value k, then all the elements of the matrix are multiplied with this scalar constant k, however, the order of matrix kA remain same, i.e., m × n.

Now, the order of the transpose of the matrix kA, i.e., (kA)t will be n × m.

As the order of the matrix A is m × n, the order of its transpose matrix, i.e., At will be n × m.

If matrix At is multiplied by the scalar value k, then the order of the matrix kAt will also be n × m.

So, the order of the matrices (kA)t and kAt is the same, i.e., n × m.

Now, let us prove that the corresponding elements of (kA)t and kAt are equal.

The (i, j)th element of (kA)t will be equal to the (j, i)th element of kA.

(i, j)th element of (kA)t = (j, i)th element of kA

⇒ (i, j)th element of (kA)t = (i, j)th element of kAt

So, we say that the corresponding elements of (kA)t and kAt are equal.

As the order and corresponding elements of (kA)t and kAt are equal, 

Therefore, we can conclude that (kA)t = kAt.

  • Transpose of Addition of Matrices: This property says that. 

(A + B)t = At + Bt

Proof:

Here A and B are two matrices of order m × n

Let, A = [a(ij)] and B = [b(ij)] of order m × n.

So, (A + B) is also of order m × n matrix

Also, At and Bt are of order n × m matrices.

So, the Transpose of matrix (A + B) or (A + B)t is an n × m matrix.

Now we can say, At + Bt is also an n × m matrix.

Now, from the transpose rule,
(j, i)th element of (A + B)t = (i, j)th element of (A + B)

= (i, j)th element of A + (i, j)th element of B 
= (j, i)th element of At + (j, i)th element of Bt
= (j, i)th element of (At + Bt)

Therefore,

(A + B)t = At + Bt

  • If “A” is a square matrix of any order and is invertible, then the inverse of its transpose is equal to the transpose of the inverse of the original matrix, i.e., (At)-1 = (A-1)t.

Proof:

To prove that (At)-1 = (A-1)t, let us consider a non-singular square matrix A.

RHS = (A-1)t

Now, multiply (A-1)t by At

= (A-1)t × At

We know that (AB)t = BtAt

So, (A-1)tAt = (AA-1)t

We know that the AA-1 = I, where “I” is an identity matrix.

So, (A-1)tAt = It

⇒ (A-1)tAt = I (Since, It = I)

⇒ (A-1)t = (At)-1 = LHS

Hence proved.

Therefore, (At)-1 = (A-1)t

Also, Check

Solved Examples on Transpose of a Matrix

Example 1: Find the transpose of the matrix A =  \begin{bmatrix} a & b & c\\ p & q & r \end{bmatrix}

Solution:

The transpose of matrix A is At 

At\begin{bmatrix} a & p \\ b & q \\ c & r \end{bmatrix}

Example 2: For matrices, A =  \begin{bmatrix} -2 & 1 & 3\\ 0 & 4 & -1 \end{bmatrix}      and  B = \begin{bmatrix} 2 & 1 \\-3 & 0 \\ 4 & -5 \end{bmatrix}

Prove that for these matrices hold the property, (AB)t = (Bt)(At)

Solution:

Here A and B are 2 × 3 and 3 × 2 matrices respectively. So, by the product rule of a matrix, we can find their product and the final matrices would be of 2 × 2 matrix.

L.H.S

Now,

AB=  \begin{bmatrix} -2 & 1 & 3\\ 0 & 4 & -1 \end{bmatrix} \times  \begin{bmatrix} 2 & 1 \\-3 & 0 \\ 4 & -5 \end{bmatrix}

 AB =\begin{bmatrix} (-2)×2+1×(-3)+3×4 & (-2)×1+1×0+3×(-5) \\ 0×2+4×(-3)+(-1)×4 & 0×1+4×0+(-1)×(-5) \end{bmatrix}

AB= \begin{bmatrix} 5 & -17 \\ -16 & 5 \end{bmatrix}

So, Transpose of matrix AB is,

(AB)^{t} =  \begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}
\begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}

R.H.S

A^{t} =  \begin{bmatrix} -2 & 0 \\ 1 & 4 \\ 3 & -1 \end{bmatrix}

and

B^{t} = \begin{bmatrix} 2 & -3 & 4 \\ 1 & 0 & -5 \end{bmatrix}

So,

B^{t}A^{t} =  \begin{bmatrix} 2 & -3 & 4\\ 1 & 0 & -5 \end{bmatrix} \times  \begin{bmatrix} -2 & 0 \\ 1 & 4 \\ 3 & -1 \end{bmatrix}\\ B^{t}A^{t} =  \begin{bmatrix} 2×(-2)+(-3)×1+4×3 & 2×0+(-3)×4+4×(-1) \\ 1×(-2)+0×1+(-5)×3 & 1×0+0×4+(-5)×(-1) \end{bmatrix}\\ B^{t}A^{t} =  \begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}

Therefore,

(AB)t = BtAt

Example 3: Verify whether (QT)T = Q or not.

Q = \left[\begin{array}{cc} 1 & 5\\ 2 & 6\\ 3 & 8 \end{array}\right]

Solution:

Q = \left[\begin{array}{cc} 1 & 5\\ 2 & 6\\ 3 & 8 \end{array}\right]

Q^{T} = \left[\begin{array}{cc} 1 & 5\\ 2 & 6\\ 3 & 8 \end{array}\right]^{T} = \left[\begin{array}{ccc} 1 & 2 & 3\\ 5 & 6 & 8 \end{array}\right]

(Q^{T})^{T} = \left[\begin{array}{ccc} 1 & 2 & 3\\ 5 & 6 & 8 \end{array}\right]^{T}

(Q^{T})^{T} = \left[\begin{array}{cc} 1 & 5\\ 2 & 6\\ 3 & 8 \end{array}\right] = Q

Hence verified.

Example 4: Verify whether the matrix given below is symmetric or not.

P = \left[\begin{array}{cc} 6 & -5\\ -5 & 6 \end{array}\right]

Solution:

We know that a square matrix “P” of order “n × n” is said to be a symmetric matrix if its transpose is the same as the original matrix, i.e., PT = P.

P^{T} = \left[\begin{array}{cc} 6 & -5\\ -5 & 6 \end{array}\right]^{T}

Now, PT is obtained by interchanging its rows into columns.

P^{T} = \left[\begin{array}{cc} 6 & -5\\ -5 & 6 \end{array}\right] = P

As PT = P, the given square matrix is symmetric.

Example 5: For matrices A= \begin{bmatrix} -1 & 5 \\ 3 & 2 \end{bmatrix}      and B= \begin{bmatrix} 3 & -2 \\5 & 4 \end{bmatrix}

Prove that these matrices hold this property, (A + B)t = At + Bt

Solution:

L.H.S

(A+B)= \begin{bmatrix} -1 & 5 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\5 & 4 \end{bmatrix}
= \begin{bmatrix} (-1)+3 & 5+(-2) \\ 3+5 & 2+4 \end{bmatrix}
= \begin{bmatrix} 2 & 3 \\ 8 & 6 \end{bmatrix}

So, 

(A+B)^{t} =  \begin{bmatrix} 2 & 8 \\ 3 & 6 \end{bmatrix}

R.H.S

A^{t} =  \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix}
 \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix}

 and, 

B^{t} =  \begin{bmatrix} 3 & 5 \\ -2 & 4 \end{bmatrix}

Now, 

A^{t} + B^{t} =  \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix} +  \begin{bmatrix} 3 & 5 \\ -2 & 4 \end{bmatrix}
= \begin{bmatrix} (-1)+3 & 3+5 \\ 5+(-2) & 2+4 \end{bmatrix}
A^{t} + B^{t} =  \begin{bmatrix} 2 & 8 \\ 3 & 6 \end{bmatrix}

Therefore, 

(A + B)t = At + Bt

FAQs on Transpose of a Matrix

Question 1: What is the Transpose of a Matrix?

Answer:

Transpose of a matrix is a matrix that is obtained by interchanging the rows and columns of the matrix. The transpose of matrix A is denoted as At. For a given matrix of order m×n, transpose of matrix is of order n×m.

Question 2: What is the order of the Transpose of a Square Matrix?

Answer:

For a square matrix order of matrix does not change on transpoe, therefore for a matrix of order n×n, the order of its transpose is also n×n. 

Question 3: What is the Addition Property of the Transpose Matrix?

Answer:

Addition property of transpose of matrix states that the sum of two transpose matrices is always equal to the sum of the transpose of individual matrices, i.e.,

(A+B)′ = A′+B′

Question 4: What is the Multiplication Property of the Transpose Matrix?

Answer:

Multiplication property of transpose of matrix states that the Product of the transpose of two matrices is always equal to the product of the transpose of individual matrices in reverse order, i.e.,

(A×B)′ = B′ × A′


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Last Updated : 13 Mar, 2023
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