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

An amount of Rs 10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined income is Rs 1310 and the combined income of first and  second investment is Rs 190 short of the income from the third. Find the investment in each using matrix method.

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.

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