Before we compute the product of $\displaystyle A$, and $\displaystyle B$, we need to check if it is possible to take the product. We will check the dimensions. $\displaystyle A$ is $\displaystyle 3\times1$, and $\displaystyle B$ is $\displaystyle 1\times3$, so the dimensions of the resulting matrix will be $\displaystyle 3\times3$. Now let's compute it.

$\displaystyle A\cdot B=\begin{bmatrix} 2\cdot1 & 2\cdot3 & 2\cdot 5\\ 3 \cdot 1& 3\cdot3 & 3 \cdot 5 \\ 4 \cdot 1 & 4\cdot3 & 4\cdot 5 \end{bmatrix} =\begin{bmatrix} 2 & 6& 10\\ 3& 9& 15 \\ 4& 12& 20 \end{bmatrix}$

$\displaystyle A\cdot B=1\cdot3+2\cdot5=3+10=13$

By definition, 

$\displaystyle \textbf{A} \times \textbf{B} = \begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 4& -2& 0\\ 1& 3& 0 \end{vmatrix} =\hat{i}(0)-\hat{j}(0)+\hat{k}[4(3)-1(-2)] = 14 \hat{k}$.

By definition, the resultant cross product vector (in this case, $\displaystyle \bf{C}$) is orthogonal to the original vectors that were crossed (in this case, $\displaystyle \bf{A}$ and $\displaystyle \bf{B}$).  In $\displaystyle 3D$, this means that $\displaystyle \bf{C}$ is a vector that is normal to the plane containing $\displaystyle \bf{A}$ and $\displaystyle \bf{B}$.

$\displaystyle \begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }1\times6\\&\text{and B has dimensions: }6\times1\\&\text{The two vectors can be multiplied:}\\&\begin{bmatrix}-18&12&-11&4&-19&13\end{bmatrix}\times\begin{bmatrix}-6\\-2\\-17\\-20\\-8\\10\end{bmatrix}\\&(-18)(-6)+(12)(-2)+(-11)(-17)+(4)(-20)+(-19)(-8)+(13)(10)\\&473\end{align*}$

$\displaystyle \begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our vectors have dimensions: }1\times2\text{ and }2\times1\\&\text{The two can be multiplied to find a product:}\\&\begin{bmatrix}-7&-6\end{bmatrix}\times\begin{bmatrix}11\\-9\end{bmatrix}\\&(-7)(11)+(-6)(-9)\\&-23\end{align*}$

$\displaystyle \begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }1\times6\\&\text{and B has dimensions: }6\times1\\&\text{The two vectors can be multiplied:}\\&\begin{bmatrix}18&-15&10&-15&-14&11\end{bmatrix}\times\begin{bmatrix}6\\-5\\-13\\9\\10\\-8\end{bmatrix}\\&(18)(6)+(-15)(-5)+(10)(-13)+(-15)(9)+(-14)(10)+(11)(-8)\\&-310\end{align*}$

$\displaystyle \begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }1\times4\\&\text{and B has dimensions: }4\times1\\&\text{It is possible to find a value for }C:\\&C=\begin{bmatrix}16&-1&6&-20\end{bmatrix}\times\begin{bmatrix}10\\19\\-19\\1\end{bmatrix}\\&C=(16)(10)+(-1)(19)+(6)(-19)+(-20)(1)\\&C=7\end{align*}$

$\displaystyle \begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }1\times2\\&\text{and B has dimensions: }2\times1\\&\text{The two vectors can be multiplied:}\\&\begin{bmatrix}13&8\end{bmatrix}\times\begin{bmatrix}-12\\7\end{bmatrix}\\&(13)(-12)+(8)(7)\\&-100\end{align*}$

$\displaystyle \begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our vectors have dimensions: }1\times6\text{ and }6\times1\\&\text{The two can be multiplied to find a product:}\\&\begin{bmatrix}-11&19&13&15&-18&6\end{bmatrix}\times\begin{bmatrix}3\\-14\\9\\-15\\-16\\10\end{bmatrix}\\&(-11)(3)+(19)(-14)+(13)(9)+(15)(-15)+(-18)(-16)+(6)(10)\\&-59\end{align*}$