Q.Find the transpose of $A=\begin{bmatrix}3&-1&2\\0&4&5\end{bmatrix}$.

Transpose the first equation and combine it with the second to eliminate $B^T$, then solve for $A$.

Let $M_1 = \begin{pmatrix}2&5\\10&2\end{pmatrix}$ and $M_2=\begin{pmatrix}1&8\\4&1\end{pmatrix}$.

Given: $2A^T+B = M_1$ ... (i), and $2B^T+A=M_2$ ... (ii).

Transpose (i) (using $(A^T)^T=A$): $2A+B^T = M_1^T$ ... (iii), where $M_1^T = \begin{pmatrix}2&10\\5&2\end{pmatrix}$.

From (ii): $B^T = \dfrac{M_2-A}{2}$. Substitute into (iii):

$2A + \frac{M_2-A}{2} = M_1^T$

Multiply through by $2$: $4A + M_2 - A = 2M_1^T \Rightarrow 3A = 2M_1^T - M_2$.

Now $2M_1^T = \begin{pmatrix}4&20\\10&4\end{pmatrix}$, so $2M_1^T - M_2 = \begin{pmatrix}4-1&20-8\\10-4&4-1\end{pmatrix} = \begin{pmatrix}3&12\\6&3\end{pmatrix}$.

Dividing by $3$: $A = \begin{pmatrix}1&4\\2&1\end{pmatrix}$.

(Check: $A^T=\begin{pmatrix}1&2\\4&1\end{pmatrix}$, so $B = M_1-2A^T = \begin{pmatrix}0&1\\2&0\end{pmatrix}$, and $2B^T+A = \begin{pmatrix}0&4\\2&0\end{pmatrix}+\begin{pmatrix}1&4\\2&1\end{pmatrix}=\begin{pmatrix}1&8\\4&1\end{pmatrix}=M_2$ — matches.)