What is Transpose of Matrix – Definition and Example

Here you will learn what is transpose of matrix with definition and examples.

Let’s begin –

What is Transpose of Matrix

Let A = \([a_{ij}]\) be a \(m\times n\) matrix. Then the transpose of A, denoted by \(A^{T}\) or A’, is an \(n\times m\) matrix such that

\((A^T)_{ij}\) = \(a_{ji}\) for all i = 1, 2, ….. , m;  j = 1, 2, ….., n.

Thus, \(A^{T}\) is obtained from A by changing its rows into columns and columns into rows.

for example, if A = \(\begin{bmatrix} 1 & 2 & 3 & 4\\ 2 &  3 & 4 & 1\\ 3 & 2 & 1 & 4  \end{bmatrix}\),

then \(A^T\) = \(\begin{bmatrix} 1 & 2 & 3 \\ 2 &  3 & 2 \\ 3 & 4 &  1 \\ 4 & 1 & 4  \end{bmatrix}\)

The first row of \(A^T\)  is the first column of A. The second row of \(A^T\) is the second column of A and so on.

Properties of Transpose

(a) for any matrix A, \((A^T)^T\) = A.

(b) for any two matrices A and B of the same order, \((A + B)^T)\) = \(A^T\) + \(B^T\).

(c) If A is a matrix and k is a scalar, then \((kA)^T\) = k\((A^T)\).

(d) If A and B are two matrices such that AB is defined, then \((AB)^T\) = \(B^T\)\(A^T\).

Generalisation : If A, B, C are three matrices confirmable for the products (AB)C and A(BC), then \((ABC)^T\) = \(C^T\)\(B^T\)\(A^T\).

The above law is called reversal law for transposes i.e. the transpose of the product is the product of the transposes taken in the reverse order.

Example : If A = \(\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}\) and B = \(\begin{bmatrix} -2 & -1 & -4 \end{bmatrix}\), verify that \((AB)^T\) = \(B^T\) \(A^T\)

Solution : We have,

A = \(\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}\) and B = \(\begin{bmatrix} -2 & -1 & -4 \end{bmatrix}\)

\(\therefore\) AB = \(\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}\) \(\begin{bmatrix} -2 & -1 & -4 \end{bmatrix}\) = \(\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12  \end{bmatrix}\)

\(\implies\) \((AB)^T\) = \(\begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12  \end{bmatrix}\)

Also, \(B^T\) = \(\begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}\) and \(A^T\) = \(\begin{bmatrix} -1 & 2 & 3 \end{bmatrix}\)

\(B^T\)\(A^T\) = \(\begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}\) \(\begin{bmatrix} -1 & 2 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12  \end{bmatrix}\)

Hence \((AB)^T\) = \(B^T\) \(A^T\)

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