If \[A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  2&0
\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  1&{12}
\end{array}} \right]\]then
A.\[AB = 0,BA = 0\]
B.\[AB = 0,BA \ne 0\]
C.\[AB \ne 0,BA = 0\]
D.\[AB \ne 0,BA \ne 0\]

Hint: For solving this problem we should know how to multiply two matrices. The first basic and important rule of multiplying two matrices is to check out the rows and columns. If the number of columns in the first matrix is equal to the number of rows in the second matrix, only then we can multiply the two matrices.

Formula used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]

Complete step by step Solution:
We are given that
\[A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  2&0
\end{array}} \right]\]
A has 2 rows and 2 columns.
\[B = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  1&{12}
\end{array}} \right]\]
B has 2 rows and 2 columns.
Since the number of columns of matrix A is equal to the number of rows of matrix B, we can multiply A and B.
\[AB = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  2&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  0&0 \\
  1&{12}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right] = 0\]
\[BA = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  1&{12}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  2&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  {25}&0
\end{array}} \right] \ne 0\]

Therefore, the correct option is (B).

Note:Students must have knowledge of matrix multiplication to solve this question. Also, one must note that matrix multiplication is not always commutative i.e., matrix AB may not always be equal to matrix BA.