Q.Construct an $m \times n$ matrix $A = [a_{ij}]$, where $a_{ij}$ is given by

To construct $A=[a_{ij}]_{m\times n}$ we simply compute $a_{ij}$ for $i=1,\dots,m$ and $j=1,\dots,n$, in that order, and place the result in row $i$, column $j$.

Step 1. Part (i): set up the $2\times 3$ grid. Here $a_{ij}=\dfrac{(i-2j)^2}{2}$, $i=1,2$ and $j=1,2,3$.

Step 2. Part (i): compute each entry.

$a_{11}=\dfrac{(1-2)^2}{2}=\dfrac12,\quad a_{12}=\dfrac{(1-4)^2}{2}=\dfrac92,\quad a_{13}=\dfrac{(1-6)^2}{2}=\dfrac{25}2$

$a_{21}=\dfrac{(2-2)^2}{2}=0,\quad a_{22}=\dfrac{(2-4)^2}{2}=2,\quad a_{23}=\dfrac{(2-6)^2}{2}=8$

So $A=\begin{pmatrix} \frac12 & \frac92 & \frac{25}2 \\ 0 & 2 & 8\end{pmatrix}$.

Step 3. Part (ii): set up the $3\times4$ grid. Here $a_{ij}=\dfrac{|3i-4j|}{4}$, $i=1,2,3$ and $j=1,2,3,4$.

Step 4. Part (ii): compute each entry.

Row $i=1$: $a_{11}=\frac{|3-4|}{4}=\frac14,\ a_{12}=\frac{|3-8|}{4}=\frac54,\ a_{13}=\frac{|3-12|}{4}=\frac94,\ a_{14}=\frac{|3-16|}{4}=\frac{13}4$

Row $i=2$: $a_{21}=\frac{|6-4|}{4}=\frac12,\ a_{22}=\frac{|6-8|}{4}=\frac12,\ a_{23}=\frac{|6-12|}{4}=\frac32,\ a_{24}=\frac{|6-16|}{4}=\frac52$

Row $i=3$: $a_{31}=\frac{|9-4|}{4}=\frac54,\ a_{32}=\frac{|9-8|}{4}=\frac14,\ a_{33}=\frac{|9-12|}{4}=\frac34,\ a_{34}=\frac{|9-16|}{4}=\frac74$

So $A=\begin{pmatrix} \frac14 & \frac54 & \frac94 & \frac{13}4 \\ \frac12 & \frac12 & \frac32 & \frac52 \\ \frac54 & \frac14 & \frac34 & \frac74\end{pmatrix}$.

• Mixing up the row index $i$ and column index $j$ when substituting
• Forgetting the outer $|\cdot|$ or the $/2, /4$ scaling factor