Q.A matrix $A=[a_{ij}]$ of order $2\times 3$ is defined by $a_{ij}=i+j$. Write down the matrix $A$ and state its order and the number of elements it has.

[!TLDR] Substitute $i=1,2$ and $j=1,2,3$ into $a_{ij}=i+j$. [!ANSWER] $A=\begin{bmatrix}2&3&4\\3&4&5\end{bmatrix}$, order $2\times3$, with $6$ elements.

Since $A$ has order $2\times3$, $i$ runs over $1,2$ and $j$ runs over $1,2,3$. Computing each entry from $a_{ij}=i+j$: $a_{11}=1+1=2$, $a_{12}=1+2=3$, $a_{13}=1+3=4$; $a_{21}=2+1=3$, $a_{22}=2+2=4$, $a_{23}=2+3=5$. Assembling these gives $A=\begin{bmatrix}2&3&4\\3&4&5\end{bmatrix}$. Since $A$ has $2$ rows and $3$ columns, its order is $2\times3$ and it has $2\times3=6$ elements. [!ANSWER] $A=\begin{bmatrix}2&3&4\\3&4&5\end{bmatrix}$, order $2\times3$, $6$ elements.

Fix the row index $i$, let the column index $j$ run through its full range substituting into the formula, then move to the next row.

A common slip is swapping the roles of $i$ and $j$ (treating the first index as the column), which for this particular symmetric-looking formula happens not to change the answer, but would silently give a wrong matrix for a formula like $a_{ij}=2i-j$.