# Construct a 3 × 4 matrix A = [aij] whose elements aij are given by:

Question:

Construct a $3 \times 4$ matrix $A=\left[a_{i j}\right]$ whose elements $a_{i j}$ are given by:

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

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

(iii) $a_{i j}=2 i$

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

(v) $a_{i j}=\frac{1}{2}|-3 i+j|$

Solution:

(i)

$a_{i j}=i+j$

Here,

$a_{11}=1+1=2, a_{12}=1+2=3, a_{13}=1+3=4, a_{14}=1+4=5$

$a_{21}=2+1=3, a_{22}=2+2=4, a_{23}=2+3=5, a_{24}=2+4=6$

$a_{31}=3+1=4, a_{32}=3+2=5, a_{33}=3+3=6$ and $a_{34}=3+4=7$

So, the required matrix is $\left[\begin{array}{cccc}2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7\end{array}\right]$.

(ii)

$a_{i j}=i-j$

Here,

$a_{11}=1-1=0, a_{12}=1-2=-1, a_{13}=1-3=-2, a_{14}=1-4=-3$

$a_{21}=2-1=1, a_{22}=2-2=0, a_{23}=2-3=-1, a_{24}=2-4=-2$

$a_{31}=3-1=2, a_{32}=3-2=1, a_{33}=3-3=0$ and $a_{34}=3-4=-1$

So, the required matrix is $\left[\begin{array}{cccc}0 & -1 & -2 & -3 \\ 1 & 0 & -1 & -2 \\ 2 & 1 & 0 & -1\end{array}\right]$.

$(i i i)$

$a_{i j}=2 i$

Here,

$a_{11}=2(1)=2, a_{12}=2(1)=2, a_{13}=2(1)=2, a_{14}=2(1)=2$

$a_{21}=2(2)=4, a_{22}=2(2)=4, a_{23}=2(2)=4, a_{24}=2(2)=4$

$a_{31}=2(3)=6, a_{32}=2(3)=6, a_{33}=2(3)=6$ and $a_{34}=2(3)=6$

So, the required matrix is $\left[\begin{array}{llll}2 & 2 & 2 & 2 \\ 4 & 4 & 4 & 4 \\ 6 & 6 & 6 & 6\end{array}\right]$.

(iv)

$a_{i j}=j$

Here,

$a_{11}=1, a_{12}=2, a_{13}=3, a_{14}=4$

$a_{21}=1, a_{22}=2, a_{23}=3, a_{24}=4$

$a_{31}=1, a_{32}=2, a_{33}=3$ and $a_{34}=4$

So, the required matrix is $\left[\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4\end{array}\right]$.

$(v)$

$a_{i j}=\frac{1}{2}|-3 i+j|$

Here,

$a_{11}=\frac{1}{2}|-3(1)+1|=\frac{1}{2}|-2|=1, a_{12}=\frac{1}{2}|-3(1)+2|=\frac{1}{2}|-1|=\frac{1}{2}, a_{13}=\frac{1}{2}|-3(1)+3|=\frac{1}{2}|0|=\frac{0}{2}=0, a_{14}=\frac{1}{2}|-3(1)+4|=\frac{1}{2}|1|=\frac{1}{2}$

$a_{21}=\frac{1}{2}|-3(2)+1|=\frac{1}{2}|-5|=\frac{5}{2}, a_{22}=\frac{1}{2}|-3(2)+2|=\frac{1}{2}|-4|=2, a_{23}=\frac{1}{2}|-3(2)+3|=\frac{1}{2}|-3|=\frac{3}{2}, a_{24}=\frac{1}{2}|-3(2)+4|=\frac{1}{2}|-2|=1$

$a_{31}=\frac{1}{2}|-3(3)+1|=\frac{1}{2}|-8|=4, a_{32}=\frac{1}{2}|-3(3)+2|=\frac{1}{2}|-7|=\frac{7}{2}, a_{33}=\frac{1}{2}|-3(3)+3|=\frac{1}{2}|-6|=3$ and $a_{34}=\frac{1}{2}|-3(3)+4|=\frac{1}{2}|-5|=\frac{5}{2}$

So, the required matrix is $\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}\right]$.