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