# Construct a 2 × 3 matrix

Question:

Construct a $2 \times 3$ matrix whose elements $a_{i j}$ are given by :

(i) $a_{i j}=i . j$

(ii) $a_{i j}=2 j-j$

(iii) $a_{i j}=i+j$

(iv) $\mathrm{a}_{i j}=\frac{(i+j)^{2}}{2}$

Solution:

(i)

Here,

$a_{i j}=i \cdot j, 1 \leq i \leq 2$ and $1 \leq j \leq 3$

$a_{11}=1 \times 1=1, a_{12}=1 \times 2=2, a_{13}=1 \times 3=3$

$a_{21}=2 \times 1=2, a_{22}=2 \times 2=4$ and $a_{23}=2 \times 3=6$

Required matrix $=A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6\end{array}\right]$

$(i i)$

Here,

$a_{i j}=2 i-j$

$a_{11}=2(1)-1=2-1=1, a_{12}=2(1)-2=2-2=0, a_{13}=2(1)-3=2-3=-1$

$a_{21}=2(2)-1=4-1=3, a_{22}=2(2)-2=4-2=2$ and $a_{23}=2(2)-3=4-3=1$

Required matrix $=A=\left[\begin{array}{ccc}1 & 0 & -1 \\ 3 & 2 & 1\end{array}\right]$

$(i v)$

Here,

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

$a_{11}=\frac{(1+1)^{2}}{2}=\frac{(2)^{2}}{2}=\frac{4}{2}=2, a_{12}=\frac{(1+2)^{2}}{2}=\frac{(3)^{2}}{2}=\frac{9}{2}, a_{13}=\frac{(1+3)^{2}}{2}=\frac{(4)^{2}}{2}=\frac{16}{2}=8$

$a_{21}=\frac{(2+1)^{2}}{2}=\frac{(3)^{2}}{2}=\frac{9}{2}, a_{22}=\frac{(2+2)^{2}}{2}=\frac{(4)^{2}}{2}=\frac{16}{2}=8$ and $a_{23}=\frac{(2+3)^{2}}{2}=\frac{(5)^{2}}{2}=\frac{25}{2}$

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