Let the cost of onions, wheat and rice per kg be $R s$ $X$, $R s$ $Y$ and $R s$ $Z$ respectively.

Then, the given situation can be represented by a system of equations as:

$4 x+3 y+2 z=60$

$2 x+4 y+6 z=90$

$6 x+2 y+3 z=70$

This system of equations can be written in the form of $A X=B$, where

$A=\left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right], X\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]$

$|A|=4(12-12)-3(6-36)+2(4-24)=0+90-40=50 \neq 0$

Now,

$A_{11}=0, A_{12}=30, A_{13}=-20$

$A_{21}=-5, A_{22}=0, A_{23}=10$

$A_{31}=10, A_{32}=-20, A_{33}=10$

$\therefore a d j A=\left[\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]$

$\therefore A^{-1}=\begin{array}{c}1 \\ |A|\end{array} a d j A=\frac{1}{50}\left[\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]$

Now,

$X=A^{-1} B$

$\Rightarrow X=\frac{1}{50}\left[\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]$

$\Rightarrow\left[\begin{array}{c}x \\ y \\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{c}0-450+700 \\ 1800+0-1400 \\ -1200+900+700\end{array}\right]$

$=\frac{1}{50}\left[\begin{array}{c}250 \\ 400 \\ 400\end{array}\right]$

$=\left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$

$\therefore x=5, y=8,$ and $z=8$

Hence, the cost of onions is $R s$ $5$ per $kg$, the cost of wheat is $R s$ $8$ per $kg$, and the cost of rice is $Rs$ $8$ per $kg$.