Question

Construct $2\times 2$ matrix,$A=\left[a_{ij}\right]$ whose elements are given by
$\left(i\right)a_{ij}=\frac{(i+j{)}^2}{2}$

$\left(ii\right)a_{ij}=\frac{\left(i\right)}{j}$

$\left(iii\right)a_{ij}=\frac{(i+2j{)}^2}{2}$

Answer 1

The order of the given matrix is $2\times 2$, so, $A={\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]}_{2\times 2}$ where $a_{ij}=\frac{(i+j{)}^2}{2}$. To find $a_{11}$, put i=1 and j=1
$\therefore ,a_{11}=\frac{(1+1{)}^2}{2}=2Similarly,a_{12}=\frac{(1+2{)}^2}{2}=\frac{9}{2}.a_{21}=\frac{(2+1{)}^2}{2}=\frac{9}{2}and{a}_{22}=\frac{(2+2{)}^2}{2}=8$
Hence, the required matrix is $A={\left[\begin{array}{cc}2& \frac{9}{2}\\ \frac{9}{2}& 8\end{array}\right]}_{2\times 2}$

Here, $A={\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]}_{2\times 2}$, where $a_{ij}=\frac{i}{j}.$
$\therefore a_{11}=\frac{1}{1}=1,a_{12}=\frac{1}{2},a_{21}=\frac{2}{1}=2and{a}_{22}=\frac{2}{2}=1$
Hence, the required matrix is $A={\left[\begin{array}{cc}2& \frac{1}{2}\\ 2& 1\end{array}\right]}_{2\times 2}$

Here, $A={\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]}_{2\times 2}$, where $a_{ij}=\frac{(i+2j{)}^2}{2}$
$\therefore a_{11}=\frac{(1+2{)}^2}{2}=\frac{9}{2},a_{12}=\frac{(1+4{)}^2}{2}=\frac{25}{2},a_{21}=\frac{(2+2{)}^2}{2}=8and{a}_{22}=\frac{(2+4{)}^2}{2}=18$
Hence, the required matrix is $A={\left[\begin{array}{cc}\frac{9}{2}& \frac{25}{2}\\ 8& 8\end{array}\right]}_{2\times 2}$