Question

For \[2 \times 3\] matrix \[A = \left[ {{a_{ij}}} \right]\] whose elements are given by \[{a_{ij}} = \dfrac{{{{(i + j)}^2}}}{2}\] then A is equal to
A. \[\left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  2&8&{\dfrac{9}{2}}
\end{array} \\
  \begin{array}{*{20}{c}}
  8&{\dfrac{9}{2}}&{\dfrac{{25}}{2}}
\end{array} \\
\end{gathered} \right]\]
B. \[\left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  2&{\dfrac{9}{2}}&8
\end{array} \\
  \begin{array}{*{20}{c}}
  {\dfrac{9}{2}}&8&{\dfrac{{25}}{2}}
\end{array} \\
\end{gathered} \right]\]
C. \[\left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  2&{\dfrac{9}{2}}&8
\end{array} \\
  \begin{array}{*{20}{c}}
  8&{\dfrac{9}{2}}&{\dfrac{{25}}{2}}
\end{array} \\
\end{gathered} \right]\]
D. \[\left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  2&{\dfrac{{25}}{2}}&8
\end{array} \\
  \begin{array}{*{20}{c}}
  {\dfrac{9}{2}}&{\dfrac{9}{2}}&8
\end{array} \\
\end{gathered} \right]\]

Answer 1

Hint: We use the general method of forming a matrix using the fact that ‘i’ represents the row and ‘j’ represents the column of the matrix. An element \[{a_{ij}}\]is calculated by substituting values of respective row and column into the formula. From a matrix with calculated values and check which option matches the matrix.
* A matrix of order \[m \times n\]tells us there are m rows and n columns for the matrix.

Step-By-Step answer:
We are given a matrix A of order \[2 \times 3\]
For a matrix A, with 2 rows and 3 columns we can write \[A = \left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  {{A_{11}}}&{{A_{12}}}&{{A_{13}}}
\end{array} \\
  \begin{array}{*{20}{c}}
  {{A_{21}}}&{{A_{22}}}&{{A_{23}}}
\end{array} \\
\end{gathered} \right]\]
We know ‘i’ represents the row and ‘j’ represents the column of the matrix
Also, \[{a_{ij}} = \dfrac{{{{(i + j)}^2}}}{2}\]gives the value of each element when we substitute value of respective row and column of that element.
 Then substitute values of i and j in the matrix A
\[ \Rightarrow A = \left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  {\dfrac{{{{(1 + 1)}^2}}}{2}}&{\dfrac{{{{(1 + 2)}^2}}}{2}}&{\dfrac{{{{(1 + 3)}^2}}}{2}}
\end{array} \\
  \begin{array}{*{20}{c}}
  {\dfrac{{{{(2 + 1)}^2}}}{2}}&{\dfrac{{{{(2 + 2)}^2}}}{2}}&{\dfrac{{{{(2 + 3)}^2}}}{2}}
\end{array} \\
\end{gathered} \right]\]
Calculate the sum in numerators of the elements of the matrix
\[ \Rightarrow A = \left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  {\dfrac{{{{(2)}^2}}}{2}}&{\dfrac{{{{(3)}^2}}}{2}}&{\dfrac{{{{(4)}^2}}}{2}}
\end{array} \\
  \begin{array}{*{20}{c}}
  {\dfrac{{{{(3)}^2}}}{2}}&{\dfrac{{{{(4)}^2}}}{2}}&{\dfrac{{{{(5)}^2}}}{2}}
\end{array} \\
\end{gathered} \right]\]
Square each term in the numerators of the elements in the matrix
\[ \Rightarrow A = \left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  {\dfrac{4}{2}}&{\dfrac{9}{2}}&{\dfrac{{16}}{2}}
\end{array} \\
  \begin{array}{*{20}{c}}
  {\dfrac{9}{2}}&{\dfrac{{16}}{2}}&{\dfrac{{25}}{2}}
\end{array} \\
\end{gathered} \right]\]
Cancel the same terms or factors from the numerator and denominator of elements of the matrix wherever the cancellation is possible.
\[ \Rightarrow A = \left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  2&{\dfrac{9}{2}}&8
\end{array} \\
  \begin{array}{*{20}{c}}
  {\dfrac{9}{2}}&8&{\dfrac{{25}}{2}}
\end{array} \\
\end{gathered} \right]\]
So, matrix A is given as \[A = \left[ \begin{gathered}
  \begin{array}{*{20}{c}}
  2&{\dfrac{9}{2}}&8
\end{array} \\
  \begin{array}{*{20}{c}}
  {\dfrac{9}{2}}&8&{\dfrac{{25}}{2}}
\end{array} \\
\end{gathered} \right]\]
Since this matches the matrix in option B

\[\therefore \]Option B is correct.

Note: Many students make the mistake of leaving the values in fraction form even when they can be cancelled and converted into simpler form. Keep in mind that given options we have the simplest form and some elements as whole numbers as well, so we have to write fractions in simplest form to match the given options.