Question

if $A={\left[a_{ij}\right]}_{{}_{4\times 3}}$ where $a_{ij}=\frac{i-j}{i+j}$,then find $A$

• A. $\left[\begin{array}{ccc}0& -\frac{1}{3}& -\frac{1}{2}\\ \frac{1}{2}& 0& \frac{1}{5}\\ \frac{1}{3}& \frac{1}{5}& 0\\ \frac{3}{5}& \frac{1}{3}& \frac{1}{7}\end{array}\right]$
• B. $\left[\begin{array}{ccc}0& -\frac{1}{3}& -\frac{1}{2}\\ \frac{1}{3}& 0& -\frac{1}{5}\\ \frac{1}{2}& \frac{1}{5}& 0\\ \frac{3}{5}& \frac{1}{3}& -\frac{1}{7}\end{array}\right]$
• C. $\left[\begin{array}{ccc}0& -3& -\frac{1}{2}\\ 2& 0& 5\\ 3& 5& 0\\ \frac{3}{5}& 3& 7\end{array}\right]$
• D. $\left[\begin{array}{ccc}0& \frac{1}{3}& \frac{1}{2}\\ -\frac{1}{3}& 0& \frac{1}{5}\\ -\frac{1}{2}& -\frac{1}{5}& 0\\ -\frac{3}{5}& -\frac{1}{3}& -\frac{1}{7}\end{array}\right]$

Answer 1

Answer: (B)

Here, $a_{ij}=\frac{i-j}{i+j}$
$a_{11}=\frac{1-1}{1+1}=0$,
$a_{12}=\frac{1-2}{1+2}=\frac{-1}{3}$,
$a_{13}=\frac{1-3}{1+3}=\frac{-1}{2}$,
$a_{21}=\frac{2-1}{2+1}=\frac{1}{3}$,
$a_{22}=\frac{2-2}{2+2}=0$,
$a_{23}=\frac{2-3}{2+3}=\frac{-1}{5},$
$a_{31}=\frac{3-1}{3+1}=\frac{1}{2}$,
$a_{32}=\frac{3-2}{3+2}=\frac{1}{5}$,
$a_{33}=\frac{3-3}{3+3}=0,$
$a_{41}=\frac{4-1}{4+1}=\frac{3}{5}$,
$a_{42}=\frac{4-2}{4+2}=\frac{1}{3}$,
$a_{43}=\frac{4-3}{4+3}=\frac{1}{7}$
so, required matrix is$\left[\begin{array}{ccc}0& -\frac{1}{3}& -\frac{1}{2}\\ \frac{1}{3}& 0& -\frac{1}{5}\\ \frac{1}{2}& \frac{1}{5}& 0\\ \frac{3}{5}& \frac{1}{3}& \frac{1}{7}\end{array}\right]$