If we switch rows and columns of matrix \(\mathbf{A}\) of type \(m \times n\) we obtain transposed (or inverted) matrix \(\mathbf{A}^T\). The elements of trasnposed matrix are:
\begin{eqnarray}
(a_{ij})^T &=& (a_{ji}).
\end{eqnarray}
The original matrix \(\mathbf{A}\) of type \(m \times n\) is transformed into matrix \(\mathbf{A}^T\) of type \(n\times m\).
\begin{eqnarray}
\mathbf{A} &=& \left(\begin{matrix}a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots& \ddots & \vdots \\ a_{m1} &a_{m2} & \cdots & a_{mn}\end{matrix}\right), \quad
\mathbf{A}^T &=& \left(\begin{matrix}a_{11} & a_{21} & \cdots & a_{m1}\\ a_{12}&a_{22}& \cdots & a_{m2}\\ \vdots& \vdots & \ddots & \vdots \\ a_{1n}&a_{2n}&\cdots&a_{mn}\end{matrix}\right).
\end{eqnarray}
It should be noted:
- If we transpose the original matrix \(\mathbf{A}\) two times we will obtain the original matrix \(\mathbf{A}\).
- The transposed sum of two matrices is equal to the sum of transposed matrices. This statement can be written in mathematical form as: \begin{eqnarray} (\mathbf{A}+\mathbf{B})^T &=& \mathbf{A}^T + \mathbf{B}^T. \end{eqnarray}
- The determinant of matix \(\mathbf{A}\) is equal to determinant of transpose matrix \(\mathbf{A}^T\): \begin{eqnarray} \det \mathbf{A} &=& \det \mathbf{A}^T. \end{eqnarray}
- The transposed matrix of the product of two matrices is equal to the product of the transposed matrices taken in reverse order: \begin{eqnarray} (\mathbf{A}\mathbf{B})^T &=& \mathbf{A}^T\mathbf{B}^T. \end{eqnarray}
- Example Transpose the matrix \(\mathbf{A} = \left(\begin{matrix}2&5&7\\ 3&6&8\\ 1&2&5\end{matrix}\right)\).
Solution \begin{eqnarray} \mathbf{A}^T&=& \left(\begin{matrix}2&3&1\\ 5&7&2\\ 7&8&5\end{matrix}\right) \end{eqnarray}