If \[{a_{ij}} = \dfrac{1}{2}\left( {3i - 2j} \right)\]and \[A = {\left[ {{a_{ij}}} \right]_{2 \times 2}}\], then A is equal to
A. \[\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&2 \\
  { - \dfrac{1}{2}}&1
\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{ - \dfrac{1}{2}} \\
  2&1
\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
  2&2 \\
  {\dfrac{1}{2}}&{ - \dfrac{1}{2}}
\end{array}} \right]\]
D. None of these

Hint: Get the value of \[{\left[ {{a_{ij}}} \right]_{2 \times 2}}\] in order to solve this problem. We must enter the values of I and j in the above equation in order to obtain the values. Given that there are 2 rows and 2 columns in this matrix, the values of i will be 1 and 2 and the values of j will be 1 and 2.

Complete step by step solution:
We are given that,
\[A = {\left[ {{a_{ij}}} \right]_{2 \times 2}}\]
The total number of elements in A is 4.
\[ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right]\]
We are also given that \[{a_{ij}} = \dfrac{1}{2}\left( {3i - 2j} \right)\]
\[{a_{11}} = \dfrac{1}{2}(3\left( 1 \right) - 2\left( 1 \right)) = \dfrac{1}{2}\]
\[{a_{12}} = \dfrac{1}{2}(3\left( 1 \right) - 2\left( 2 \right)) = - \dfrac{1}{2}\]
\[{a_{21}} = \dfrac{1}{2}(3\left( 2 \right) - 2\left( 1 \right)) = 2\]
\[{a_{22}} = \dfrac{1}{2}(3\left( 2 \right) - 2\left( 2 \right)) = 1\]
On putting the values calculated above we get the matrix as
\[A = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{ - \dfrac{1}{2}} \\
  2&1
\end{array}} \right]\]

Option B. is the correct answer.

Note: You should always remember that i is the number of rows and j is the number of columns. \[{a_{ij}}\] stands for the element of the ith row and jth column. By putting the values of i and j in \[{a_{ij}}\] we get the required matrix.