Question

Construct a $3\times 4$ matrix, whose elements are given by
(i) $a_{ij}=\frac{1}{2}|-3i+j|$
(ii) $a_{ij}=2i-j$

Answer 1

In general a $3\times 4$ matrix is given by,
$A=\left[\begin{array}{c}{a}_{11}\\ a_{21}\\ a_{31}\end{array}\begin{array}{c}{a}_{12}\\ a_{22}\\ a_{32}\end{array}\begin{array}{c}{a}_{13}\\ a_{23}\\ a_{33}\end{array}\begin{array}{c}{a}_{14}\\ a_{24}\\ a_{34}\end{array}\right]$

(i)

$a_{ij}=\frac{1}{2}|-3i+j|,i=1,2,3andj=1,2,3,4$
$\therefore a_{11}=\frac{1}{2}|-3\times 1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1$

$a_{21}=\frac{1}{2}|-3\times 2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{5}{2}$

$a_{31}=\frac{1}{2}|-3\times 3+1|=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{8}{2}=4$

$a_{12}=\frac{1}{2}|-3\times 1+2|=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2}$

$a_{22}=\frac{1}{2}|-3\times 2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$

$a_{32}=\frac{1}{2}|-3\times 3+2|=\frac{1}{2}|-9+2|=\frac{7}{2}$

$a_{23}=\frac{1}{2}|-3\times 2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$

$a_{33}=\frac{1}{2}|-3\times 3+3|=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{6}{2}=3$

$a_{14}=\frac{1}{2}|-3\times 1+4|=\frac{1}{2}|-3+4|=\frac{1}{2}\left|1\right|=\frac{1}{2}$

$a_{24}=\frac{1}{2}|-3\times 2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{2}{2}=1$

$a_{34}=\frac{1}{2}|-3\times 3+4|=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{5}{2}$

Therefore, the required matrix is
$A=\left[\begin{array}{c}1\\ \frac{5}{2}\\ 4\end{array}\begin{array}{c}\frac{1}{2}\\ 2\\ \frac{3}{2}\end{array}\begin{array}{c}0\\ \frac{3}{2}\\ 3\end{array}\begin{array}{c}\frac{1}{2}\\ 1\\ \frac{5}{2}\end{array}\right]$

(ii)

$a_{ij}=2i-j,i=1,2,3andj=1,2,3,4$
$\therefore a_{11}=2\times 1-1=2-1=1$
$a_{21}=2\times 2-1=4-1=3$
$a_{31}=2\times 3-1=6-1=5$
$a_{12}=2\times 1-2=2-2=0$
$a_{22}=2\times 2-2=4-2=2$
$a_{32}=2\times 3-2=6-2=4$
$a_{13}=2\times 1-3=2-3=-1$
$a_{23}=2\times 2-3=4-3=1$
$a_{33}=2\times 3-3=6-3=3$
$a_{14}=2\times 1-4=2-4=-2$
$a_{24}=2\times 2-4=4-4=0$
$a_{34}=2\times 3-4=6-4=2$
Therefore, the required matrix is
$A=\left[\begin{array}{c}1\\ 3\\ 5\end{array}\begin{array}{c}0\\ 2\\ 4\end{array}\begin{array}{c}-1\\ 1\\ 3\end{array}\begin{array}{c}-2\\ 0\\ 2\end{array}\right]$