Question

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

Answer 1

Hint: The question is related to the matrix $2 \times 3$ you must know the basic structures of the matrix $2 \times 3$. Here ij is the number of the matrix using the equation given in the question to find the elements of the matrix like ${a_{ij}} = {a_{12}}$.
we have matrix $2 \times 3$
$\left[ {\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}
\end{array}}&{{a_{13}}} \\
  {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{22}}}
\end{array}}&{{a_{23}}}
\end{array}} \right]$
In ${a_{ij}} = {a_{11}}$

Complete step-by-step answer:
Given in the question that ${a_{ij}} = \dfrac{{{{(i + j)}^2}}}{2}$
Put the values of ij
${a_{11}} = \dfrac{{{{(1 + 1)}^2}}}{2}$
Open the brackets and do the square of the number
${a_{11}} = \dfrac{4}{2}$
Divide the numerator by denominator
${a_{11}} = 2$
Now ${a_{12}}$
${a_{12}} = \dfrac{{{{(1 + 2)}^2}}}{2}$
Open the brackets and do the square of the number
${a_{12}} = \dfrac{9}{2}$
Now \[{a_{13}}\]
\[{a_{13}} = \dfrac{{{{(1 + 3)}^2}}}{2}\]
Open the brackets and do the square of the number
\[{a_{13}} = \dfrac{{16}}{2}\]
Divide the numerator by denominator
\[{a_{13}} = 8\]
Now ${a_{21}}$
${a_{21}} = \dfrac{{{{(2 + 1)}^2}}}{2}$
Open the brackets and do the square of the number
${a_{21}} = \dfrac{9}{2}$
Now ${a_{22}}$
${a_{22}} = \dfrac{{{{(2 + 2)}^2}}}{2}$
Open the brackets and do the square of the number
${a_{22}} = \dfrac{{16}}{2}$
Divide the numerator by denominator
${a_{22}} = 8$
Now ${a_{23}}$
${a_{23}} = \dfrac{{{{(2 + 3)}^2}}}{2}$
Open the brackets and do the square of the number
${a_{23}} = \dfrac{{25}}{2}$
So now we have every element of the matrix $2 \times 3$
Matrix $2 \times 3$ is
A = $\left[ {\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  2&{\dfrac{9}{2}}
\end{array}}&8 \\
  {\begin{array}{*{20}{c}}
  {\dfrac{9}{2}}&8
\end{array}}&{\dfrac{{25}}{2}}
\end{array}} \right]$

Hence the option b is the correct option.

Note: The most important part of the question as well as in the solution is the matrix order which is given in the question $2 \times 3$. Here students mostly do the mistake in the calculation part.do the solution step by step. Here the order of the matrix represents the numbers of rows and the number of columns. The matrix of order $2 \times 3$ means it has 2 rows and 3 columns hence the result matrix has to be 2 rows and 3 columns.