Q.If $A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}$ and $B=\begin{bmatrix}0&1&-1\\2&3&5\end{bmatrix}$, find $A^T$ and $B^T$, and verify that $(A+B)^T=A^T+B^T$.

The transpose of $A=[a_{ij}]_{m\times n}$, written $A^T$, is obtained by interchanging the rows and columns of $A$: $A^T = [b_{ij}]_{n\times m}$ where $b_{ij}=a_{ji}$. So the $(i,j)$th entry of $A^T$ is the $(j,i)$th entry of $A$; a matrix's rows become its columns and vice-versa.

Basic properties (for matrices of suitable order, $k$ any scalar):

1. $(A^T)^T = A$ — transposing twice returns the original matrix.
2. $(kA)^T = kA^T$.
3. $(A+B)^T = A^T+B^T$. …