Question

Construct a $2\times 3$ matrix $A=\left[a_{ij}\right]$ whose elements are given by $a_{ij}=\frac{{(i+2j)}^2}{2}$

Answer 1

We have $\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]$

Since $a_{ij}=\frac{{(i+2j)}^2}{2}$

$a_{11}=\frac{{(1+2\times 1)}^2}{2}=\frac{{(1+2)}^2}{2}=\frac{{\left(3\right)}^2}{2}=\frac{9}{2}$

$a_{12}=\frac{{(1+2\times 2)}^2}{2}=\frac{{(1+4)}^2}{2}=\frac{{\left(5\right)}^2}{2}=\frac{25}{2}$

$a_{13}=\frac{{(1+2\times 3)}^2}{2}=\frac{{(1+6)}^2}{2}=\frac{{\left(7\right)}^2}{2}=\frac{49}{2}$

$a_{21}=\frac{{(2+2\times 1)}^2}{2}=\frac{{(2+2)}^2}{2}=\frac{{\left(4\right)}^2}{2}=\frac{16}{2}=8$

$a_{22}=\frac{{(2+2\times 2)}^2}{2}=\frac{{(2+4)}^2}{2}=\frac{{\left(6\right)}^2}{2}=\frac{36}{2}=18$

$a_{23}=\frac{{(2+2\times 3)}^2}{2}=\frac{{(2+6)}^2}{2}=\frac{{\left(8\right)}^2}{2}=\frac{64}{2}=32$

Hence the required matrix is

$A=\left[\begin{array}{ccc}\frac{9}{2}& \frac{25}{2}& \frac{49}{2}\\ 8& 18& 32\end{array}\right]$