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Given are two real matrices

$A=\begin{pmatrix} 2 & 5 & -3\\ 0 &-1 & 6 \end{pmatrix}$

$B=\begin{pmatrix} 3 & 1\\ 4 & -2\\ 0 & 6 \end{pmatrix}$

Is it possible to do the operation $3A+4B^{T}$ with these?

I tried:

$$3\begin{pmatrix} 2 & 5 & -3\\ 0 &-1 & 6 \end{pmatrix}+4 \begin{pmatrix} 3 & 4 & 0\\ 1 &-2 & 6 \end{pmatrix}$$

And now we multiply:

$$\begin{pmatrix} 6 & 15 & -9\\ 0 &-3 & 18 \end{pmatrix}+\begin{pmatrix} 12 & 16 & 0\\ 4 &-8 & 24 \end{pmatrix}$$

Add then we have:

$$\begin{pmatrix} 18 & 31 & -9\\ 4 &-11 & 42 \end{pmatrix}$$

So I say it's possible and I hope it's correct?

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    This seems correct. – 2017-01-09
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    Yes, as the trasnposed matrix defined : $[A^T]_{ij}=[A]_{ji}$. For more information you can read the see the animation - https://en.wikipedia.org/wiki/Transpose. – 2017-01-09

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Yes, but you don't even need to do the computation to answer your question. You just need to compute the dimensions of the matrices:

$B \ $ is $\ (3 \times 2) \implies B^{T} \ $ is $ \ (2 \times 3)$ which is also the dimension of $A$, so you can perform matrix addition, and since a scalar multiple of a matrix leaves the dimension unchanged you can compute your expression.