{"id":4992,"date":"2021-09-04T00:14:23","date_gmt":"2021-09-03T18:44:23","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4992"},"modified":"2022-01-16T17:01:35","modified_gmt":"2022-01-16T11:31:35","slug":"what-is-transpose-of-matrix","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-transpose-of-matrix\/","title":{"rendered":"What is Transpose of Matrix – Definition and Example"},"content":{"rendered":"
Here you will learn what is transpose of matrix with definition and examples.<\/p>\n
Let’s begin –<\/p>\n
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<\/p>\n
\n\\((A^T)_{ij}\\) = \\(a_{ji}\\) for all i = 1, 2, ….. , m; j = 1, 2, ….., n.<\/p>\n
Thus, \\(A^{T}\\) is obtained from A by changing its rows into columns and columns into rows.<\/p>\n<\/blockquote>\n
for example<\/strong>, if A = \\(\\begin{bmatrix} 1 & 2 & 3 & 4\\\\ 2 & 3 & 4 & 1\\\\ 3 & 2 & 1 & 4 \\end{bmatrix}\\),<\/p>\n
then \\(A^T\\) = \\(\\begin{bmatrix} 1 & 2 & 3 \\\\ 2 & 3 & 2 \\\\ 3 & 4 & 1 \\\\ 4 & 1 & 4 \\end{bmatrix}\\)<\/p>\n
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.<\/p>\n
Properties of Transpose<\/h2>\n
\n(a) for any matrix A, \\((A^T)^T\\) = A.<\/p>\n
(b) for any two matrices A and B of the same order, \\((A + B)^T)\\) = \\(A^T\\) + \\(B^T\\).<\/p>\n
(c) If A is a matrix and k is a scalar, then \\((kA)^T\\) = k\\((A^T)\\).<\/p>\n
(d) If A and B are two matrices such that AB is defined, then \\((AB)^T\\) = \\(B^T\\)\\(A^T\\).<\/p>\n<\/blockquote>\n
Generalisation<\/strong> : 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\\).<\/p>\n
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.<\/p>\n\n\n
Example : <\/span>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\\)<\/p>\n
Solution : <\/span>We have,
\n A = \\(\\begin{bmatrix} -1 \\\\ 2 \\\\ 3 \\end{bmatrix}\\) and B = \\(\\begin{bmatrix} -2 & -1 & -4 \\end{bmatrix}\\)
\n \\(\\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}\\)
\n\\(\\implies\\) \\((AB)^T\\) = \\(\\begin{bmatrix} 2 & -4 & -6 \\\\ 1 & -2 & -3 \\\\ 4 & -8 & -12 \\end{bmatrix}\\)
\nAlso, \\(B^T\\) = \\(\\begin{bmatrix} -2 \\\\ -1 \\\\ -4 \\end{bmatrix}\\) and \\(A^T\\) = \\(\\begin{bmatrix} -1 & 2 & 3 \\end{bmatrix}\\)
\n\\(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}\\)
\nHence \\((AB)^T\\) = \\(B^T\\) \\(A^T\\)
<\/p>\n\n\n\n