Question

Construct a $2\times 2$ matrix $A={{\left[ {{a}_{ij}} \right]}_{2\times 2}}$ , whose elements are ${{a}_{ij}}$ and ${{a}_{ij}}=\dfrac{{{\left( i+j \right)}^{2}}}{2}$ .

Answer 1

Hint: At first construct the matrix by putting ${{a}_{ij}}$, with the actual values of $i,j$ . Then calculate the numerical values by using ${{a}_{ij}}=\dfrac{{{\left( i+j \right)}^{2}}}{2}$.

Complete step-by-step answer:
When some numbers are arranged in rows and columns and are surrounded on both sides by square brackets, we call it a matrix.
Here we have to construct a $2\times 2$ matrix named $A$ .
Now, this $2\times 2$ is known as the order of the matrix. Order of matrices helps us to know how many rows and how many columns there are in a matrix.
Generally we denote the order of a matrix by $m\times n$, where $m$ denotes the number of rows and $n$ denotes the number of columns. Here the order of the matrix is $2\times 2$ . That means this matrix has 2 rows and 2 columns.
Now, the elements of the matrix are generally denoted by ${{a}_{ij}}$. Where $ij$ help us to understand the actual position of the element.
In a matrix $A$, an element in row $i$ and column $j$ is represented by ${{a}_{ij}}$.
${{a}_{11}}$ is the element in 1st row and 1st column.
${{a}_{12}}$ is the element in 1st row and 2nd column.
${{a}_{21}}$ is the element in the 2nd row and 1st column.
${{a}_{22}}$ is the element in the 2nd row and 2nd column.
Therefore, $A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right]$
Now let us substitute the values.
${{a}_{11}}=\dfrac{{{\left( 1+1 \right)}^{2}}}{2}=\dfrac{{{2}^{2}}}{2}=2$, here i is 1 and j is 1.
${{a}_{12}}=\dfrac{{{\left( 1+2 \right)}^{2}}}{2}=\dfrac{{{3}^{2}}}{2}=\dfrac{9}{2}$, here i is 1 and j is 2.
${{a}_{21}}=\dfrac{{{\left( 2+1 \right)}^{2}}}{2}=\dfrac{{{3}^{2}}}{2}=\dfrac{9}{2}$, here i is 2 and j is 1.
${{a}_{22}}=\dfrac{{{\left( 2+2 \right)}^{2}}}{2}=\dfrac{{{4}^{2}}}{2}=\dfrac{16}{2}=8$, here i is 2 and j is 2.
Therefore,
$A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right]=\left[ \begin{matrix} 2 & \dfrac{9}{2} \\\ \dfrac{9}{2} & 8 \\\ \end{matrix} \right]$
Hence, $A=\left[ \begin{matrix} 2 & \dfrac{9}{2} \\\ \dfrac{9}{2} & 8 \\\ \end{matrix} \right]$

Note: We generally make mistakes to understand the position of an element in a matrix. Always check the position of the row first then the column.