Given $A=\left[\begin{array}{c}1 \\ 0 \\ 4 \\ -2\end{array}\right]$ and $B=\left[\begin{array}{llll}6 & -7 & -1 & 8\end{array}\right]$, find $A B$ and $B A$.
\[
\begin{array}{l}
A B= \\
B A=
\end{array}
\]

Hint: Matrices need to be entered as [(elements of row 1 separated by commas), (elements of row 2 separated by commas), (elements of each row separated by commas)].

Example: $C=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]$ would be entered as $[(1,2,3),(4,5,6)]$

Step 1 ：First, let's understand the problem. We are given two matrices, A and B, and we are asked to find the product of these matrices in both orders, i.e., AB and BA.

Step 2 ：Matrix A is a column matrix with 4 rows and 1 column, and matrix B is a row matrix with 1 row and 4 columns.

Step 3 ：The product of two matrices is found by multiplying the elements of the rows of the first matrix by the corresponding elements of the columns of the second matrix and then adding these products.

Step 4 ：For AB, we have: \(AB = [(1*6 + 0*-7 + 4*-1 + -2*8)] = [(6 - 0 - 4 - 16)] = [-14]\)

Step 5 ：So, \(\boxed{AB = [-14]}\)

Step 6 ：Now, let's find BA.

Step 7 ：For BA, we have: \(BA = [(6*1, -7*1, -1*1, 8*1); (6*0, -7*0, -1*0, 8*0); (6*4, -7*4, -1*4, 8*4); (6*-2, -7*-2, -1*-2, 8*-2)] = [(6, -7, -1, 8); (0, 0, 0, 0); (24, -28, -4, 32); (-12, 14, 2, -16)]\)

Step 8 ：So, \(\boxed{BA = [(6, -7, -1, 8); (0, 0, 0, 0); (24, -28, -4, 32); (-12, 14, 2, -16)]}\)

Step 9 ：To summarize, we have: \(\boxed{AB = [-14]}\) and \(\boxed{BA = [(6, -7, -1, 8); (0, 0, 0, 0); (24, -28, -4, 32); (-12, 14, 2, -16)]}\)