Solution:

Suppose the numbers as $x$, $y$, and $z$
$x+y+z=10,000 \ldots \ldots \text { (i) }$
Also,
$0.1 \mathrm{x}+0.12 \mathrm{y}+0.15 z=1310 \ldots \ldots \text { (ii) }$
Again,
$0.1 x+0.12 y-0.15 z=-190 \ldots \ldots$ (iii) $\quad\left[\begin{array}{ccc}1 & 1 & 1 \\ 0.1 & 0.12 & 0.15 \\ 0.1 & 0.12 & -0.15\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}10000 \\ 1310 \\ -190\end{array}\right]$ $\mathrm{A} X=\mathrm{B}$ $|\mathrm{A}|=1(-0.036)-1(-0.03)+1(0)$ $=-0.006$
As a result, the unique solution given by $x=A^{-1} B$
$C_{11}=-0.036$
$C_{12}=0.27$
$C_{13}=0$
$\mathrm{C}_{21}=0.27$
$C_{22}=-0.25$
$C_{23}=-0.02$
$\mathrm{C}_{31}=0.03$
$C_{32}=-0.05$
$\mathrm{C}_{33}=0.02$
$\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A}) \mathrm{B}$
$\operatorname{Adj} A=\left[\begin{array}{ccc}-0.036 & 0.27 & 0.03 \\ 0.27 & -0.25 & -0.05 \\ 0.03 & -0.02 & 0.02\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}-0.036 & 0.27 & 0.03 \\ 0.03 & -0.25 & -0.05 \\ 0 & -0.02 & 0.02\end{array}\right]$
$\mathrm{X}=\frac{1}{-0.006}\left[\begin{array}{ccc}-0.036 & 0.27 & 0.03 \\ 0.03 & -0.25 & -0.05 \\ 0 & -0.02 & 0.02\end{array}\right]\left[\begin{array}{c}10000 \\ 1310 \\ -190\end{array}\right]$
$X=\frac{1}{-0.006}\left[\begin{array}{l}-12 \\ -18 \\ -30\end{array}\right]$
$\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
2000 \\
3000 \\
5000
\end{array}\right]$
As a result, $x=$ Rs $2000, y=\operatorname{Rs} 3000$ and $z=$ Rs 5000