$\left(A+2 B^{\top}\right)^{\top}=A^{\top}+2 B$
(ii)$\left(3 A-5 B^{\top}\right)^{\top}=3 A^{\top}-5 B$
$\therefore \quad \mathbf{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 3 & -1 \\ -3 & -2 & 4\end{array}\right]$ and $B^{\mathrm{T}}=\left[\begin{array}{cc}-3 & 2 \\ 4 & -1 \\ 1 & -3\end{array}\right]$
$\begin{aligned} \therefore \quad A+2 B^T & =\left[\begin{array}{cc}2 & -3 \\ 3 & -2 \\ -1 & 4\end{array}\right]+2\left[\begin{array}{cc}-3 & 2 \\ 4 & -1 \\ 1 & -3\end{array}\right] \\ & =\left[\begin{array}{cc}2 & -3 \\ 3 & -2 \\ -1 & 4\end{array}\right]+\left[\begin{array}{cc}-6 & 4 \\ 8 & -2 \\ 2 & -6\end{array}\right]\end{aligned}$
$\therefore \quad A+2 B^T=\left[\begin{array}{cc}-4 & 1 \\ 11 & -4 \\ 1 & -2\end{array}\right]$
$\therefore \quad\left(A+2 B^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{ccc}-4 & 11 & 1 \\ 1 & -4 & -2\end{array}\right]$
$\ldots$...(i)
$\begin{aligned} A^{\mathrm{T}}+2 B & =\left[\begin{array}{ccc}2 & 3 & -1 \\ -3 & -2 & 4\end{array}\right]+2\left[\begin{array}{ccc}-3 & 4 & 1 \\ 2 & -1 & -3\end{array}\right] \\ & =\left[\begin{array}{ccc}2 & 3 & -1 \\ -3 & -2 & 4\end{array}\right]+\left[\begin{array}{ccc}-6 & 8 & 2 \\ 4 & -2 & -6\end{array}\right] \\ & =\left[\begin{array}{ccc}-4 & 11 & 1 \\ 1 & -4 & -2\end{array}\right] \quad \ldots \text { (ii) }\end{aligned}$
From (i) and (ii), we get
$\left(A+2 B^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}+2 \mathrm{~B}$
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