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.

Answer 1

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
$\Rightarrow x+y+z=10,000$
$$\begin{gathered} \mathrm{Income}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{investment}\mathrm{of}\mathrm{Rs}x=\mathrm{Rs}\frac{10x}{100}=\mathrm{Rs}0.1x \\ \mathrm{Income}\mathrm{from}\mathrm{the}\mathrm{second}\mathrm{investment}\mathrm{of}\mathrm{Rs}x=\mathrm{Rs}\frac{12y}{100}=\mathrm{Rs}0.12y \\ \mathrm{Income}\mathrm{from}\mathrm{the}\mathrm{third}\mathrm{investment}\mathrm{of}\mathrm{Rs}x=\mathrm{Rs}\frac{15z}{100}=\mathrm{Rs}0.15z \\ \therefore \mathrm{Total}\mathrm{annual}\mathrm{income}=\mathrm{Rs}\left(0.1x+0.12y+0.15z\right) \\ \Rightarrow 0.1x+0.12y+0.15z=1310\left(\because \mathrm{Total}\mathrm{annual}\mathrm{income}=\mathrm{Rs}1310\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{combined}\mathrm{income}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{two}\mathrm{incomes}\mathrm{is}\mathrm{Rs}190\mathrm{short}\mathrm{of}\mathrm{the}\mathrm{income}\mathrm{from}\mathrm{the}\mathrm{third}. \\ \therefore 0.1x+0.12y=0.15z-190 \\ \Rightarrow -0.1x-0.12y+0.15z=190 \end{gathered}$$

$$\begin{gathered} \text{Thus, we obtain the following system of simultaneous linear equations:} \\ x+y+z=10000 \\ 0.1x+0.12y+0.15z=1310 \\ -0.1x-0.12y+0.15z=190 \\ \text{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}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}10000\\ 1310\\ 190\end{array}\right] \\ AX=B \\ \mathrm{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}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}10000\\ 1310\\ 190\end{array}\right] \\ \left|A\right|\text{=1}\left(0.15\times 0.12+0.15\times 0.12\right)-1\left(0.15\times 0.1+0.15\times 0.1\right)+1\left(-0.1\times 0.12+0.12\times 0.1\right) \\ =0.036-0.03+0 \\ =0.006 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A=}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}0.12& 0.15\\ -0.12& 0.15\end{array}\right|=0.036,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}0.1& 0.15\\ -0.1& 0.15\end{array}\right|=-0.03,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}0.1& 0.12\\ -0.1& -0.12\end{array}\right|=0 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& 1\\ -0.12& 0.15\end{array}\right|=-0.27,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 1\\ -0.1& 0.15\end{array}\right|=0.25,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 1\\ -0.1& -0.12\end{array}\right|=0.02 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& 1\\ 0.12& 0.15\end{array}\right|=0.03,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 1\\ 0.1& 0.15\end{array}\right|=-0.05,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 1\\ 0.1& 0.12\end{array}\right|=0.02 \\ \mathrm{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}{\left|A\right|}\mathrm{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}{c}x\\ y\\ z\end{array}\right]=\frac{1000}{6}\left[\begin{array}{c}12\\ 18\\ 30\end{array}\right] \\ \therefore x=2000,y=3000\mathrm{and}z=5000 \\ \mathrm{Thus},\mathrm{the}\mathrm{three}\mathrm{investments}\mathrm{are}\mathrm{of}\mathrm{Rs}2000,\mathrm{Rs}3000\mathrm{and}\mathrm{Rs}5000,\mathrm{respectively}. \end{gathered}$$