Question

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

$\left(ii\right)a_{ij}=2i-j$.

Answer 1

The order of given matrix is $3\times 4$, so the required matrix is $A={\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\end{array}\right]}_{3\times 4}$, where $a_{ij}=\frac{1}{2}|-3i+j|.$
Putting the values in place of i and j, we will find all the elements of matrix A.

$\therefore a_{11}=\frac{1}{2}|-3+1|=1,a_{12}=\frac{1}{2}|-3+2|=\frac{1}{2}{a}_{13}=\frac{1}{2}|-3+3|=0,a_{14}=\frac{1}{2}|-3+4|=\frac{1}{2}{a}_{21}=\frac{1}{2}|-6+1|=\frac{5}{2},a_{22}=\frac{1}{2}|-6+2|=2a_{23}=\frac{1}{2}|-6+3|=\frac{3}{2},a_{24}=\frac{1}{2}|-6+4|=1a_{31}=\frac{1}{2}|-9+1|=4,a_{32}=\frac{1}{2}|-9+2|=\frac{7}{2}{a}_{33}=\frac{1}{2}|-9+3|=3and{a}_{34}=\frac{1}{2}|-9+4|=\frac{5}{2}$
Hence, the required matrix is $A={\left[\begin{array}{cccc}1& \frac{1}{2}& 0& \frac{1}{2}\\ \frac{5}{2}& 2& \frac{3}{2}& 1\\ 4& \frac{7}{2}& 3& \frac{5}{2}\end{array}\right]}_{3\times 4}$

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