For the matrices A and B, verify that (AB)′= B'A' where
(i)A=\(\begin{bmatrix}1\\-4\\3\end{bmatrix},\,B=\begin{bmatrix}-1&2&1\end{bmatrix}\)
(ii)A=\(\begin{bmatrix}0\\1\\2\end{bmatrix},\,B=\begin{bmatrix}1&5&7\end{bmatrix}\)
(i)AB=\(\begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}-1&2&1\end{bmatrix}\)=\(\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}\)
therefore (AB)'=\(\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\)
Now A'=\(\begin{bmatrix}1&-4&3\end{bmatrix}\),B'=\(\begin{bmatrix}-1\\2\\1\end{bmatrix}\)
Therefore B'A'=\(\begin{bmatrix}-1\\2\\1\end{bmatrix}=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\)
Hence we verified that:(AB)′= B'A'
(ii)AB=\(\begin{bmatrix}0\\1\\2\end{bmatrix}\begin{bmatrix}1&5&7\end{bmatrix}\)=\(\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}\)
so (AB)'=\(\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)
Now A'=\(\begin{bmatrix}0&1&2\end{bmatrix}\),B'=\(\begin{bmatrix}1\\5\\7\end{bmatrix}\)
so B'A'=\(\begin{bmatrix}1\\5\\7\end{bmatrix}\)\(\begin{bmatrix}0&1&2\end{bmatrix}\)=\(\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)
Hence we verified that(AB)′= B'A'
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
The matrix acquired by interchanging the rows and columns of the parent matrix is called the Transpose matrix. The transpose matrix is also defined as - “A Matrix which is formed by transposing all the rows of a given matrix into columns and vice-versa.”
The transpose matrix of A is represented by A’. It can be better understood by the given example:
Now, in Matrix A, the number of rows was 4 and the number of columns was 3 but, on taking the transpose of A we acquired A’ having 3 rows and 4 columns. Consequently, the vertical Matrix gets converted into Horizontal Matrix.
Hence, we can say if the matrix before transposing was a vertical matrix, it will be transposed to a horizontal matrix and vice-versa.
Read More: Transpose of a Matrix