An amount of Rs 10,000 is put into three investments at the rate of 10,

Solution:

Let $x, y$ and $z$ be the investments at the rates of interest of $10 \%, 12 \%$ and $15 \%$ per annum respectively.

Total investment $=$ Rs 10,000

$\Rightarrow x+y+z=10,000$

Income from the first investment of Rs $x=\operatorname{Rs} \frac{10 x}{100}=\operatorname{Rs} 0.1 x$

Income from the second investment of Rs $x=\operatorname{Rs} \frac{12 y}{100}=$ Rs $0.12 y$

Income from the third investment of Rs $x=\operatorname{Rs} \frac{15 z}{100}=$ Rs $0.15 z$

$\therefore$ Total annual income $=\mathrm{Rs}(0.1 x+0.12 y+0.15 z)$

$\Rightarrow 0.1 x+0.12 y+0.15 z=1310 \quad(\because$ Total annual income $=$ Rs 1310$)$

It is given that the combined income from the first two incomes is Rs 190 short of the income from the third.

$\therefore 0.1 x+0.12 y=0.15 z-190$

$\Rightarrow-0.1 x-0.12 y+0.15 z=190$

Thus, we obtain the following system of simultaneous linear equations:

$x+y+z=10000$

$0.1 x+0.12 y+0.15 z=1310$

$-0.1 x-0.12 y+0.15 z=190$

The given system of equation can be written in matrix form as follows:

$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0.1 & 0.12 & 0.15 \\ -0.1 & -0.12 & 0.15\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}10000 \\ 1310 \\ 190\end{array}\right]$

$A X=B$

Here,

$A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 0.1 & 0.12 & 0.15 \\ -0.1 & -0.12 & 0.15\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}10000 \\ 1310 \\ 190\end{array}\right]$

$|A|=1(0.15 \times 0.12+0.15 \times 0.12)-1(0.15 \times 0.1+0.15 \times 0.1)+1(-0.1 \times 0.12+0.12 \times 0.1)$

$=0.036-0.03+0$

$=0.006$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}0.12 & 0.15 \\ -0.12 & 0.15\end{array}\right|=0.036, C_{12}=(-1)^{1+2}\left|\begin{array}{cc}0.1 & 0.15 \\ -0.1 & 0.15\end{array}\right|=-0.03, C_{13}=(-1)^{1+3}\left|\begin{array}{cc}0.1 & 0.12 \\ -0.1 & -0.12\end{array}\right|=0$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}1 & 1 \\ -0.12 & 0.15\end{array}\right|=-0.27, C_{22}=(-1)^{2+2}\left|\begin{array}{cc}1 & 1 \\ -0.1 & 0.15\end{array}\right|=0.25, C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & 1 \\ -0.1 & -0.12\end{array}\right|=0.02$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}1 & 1 \\ 0.12 & 0.15\end{array}\right|=0.03, C_{32}=(-1)^{3+2}\left|\begin{array}{cc}1 & 1 \\ 0.1 & 0.15\end{array}\right|=-0.05, C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & 1 \\ 0.1 & 0.12\end{array}\right|=0.02$

$\operatorname{adj} A=\left[\begin{array}{ccc}0.036 & -0.03 & 0 \\ -0.27 & 0.25 & 0.02 \\ 0.03 & -0.05 & 0.02\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{0.006}\left[\begin{array}{ccc}0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow X=\frac{1}{0.006}\left[\begin{array}{ccc}0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02\end{array}\right]\left[\begin{array}{c}10000 \\ 1310 \\ 190\end{array}\right]$

$\Rightarrow X=\frac{1}{0.006}\left[\begin{array}{c}360-353.7+5.7 \\ -300+327.5-9.5 \\ 0+26.2+3.8\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1000}{6}\left[\begin{array}{l}12 \\ 18 \\ 30\end{array}\right]$

$\therefore x=2000, y=3000$ and $z=5000$

Thus, the three investments are of Rs 2000, Rs 3000 and Rs 5000 , respectively.