Question

An amount of $Rs.5000$ is invested in three investments at rate $6\%$, $7\%$ and $8\%$ per annum respectively. The total annual income from these investments is $Rs.350$. If the total annual income from first two investments is $Rs.70$ more than the income from the third, find the amount of each investment by using determinant methods.

Answer 1

Let investments be x, y, z.

$x=\frac{6\times 5000}{100}=300$

$y=\frac{7\times 5000}{100}=350$

$z=\frac{8\times 5000}{100}=400$

Now,

$6x+7y+8z=100$ ………$\left(1\right)$

$x+y+z=358$ ……$\left(2\right)$

$x+y-z=70$ ………$\left(3\right)$

$\Delta x=B$

$\left[\begin{array}{ccc}6& 7& 8\\ 1& 1& 1\\ 1& 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}100\\ 358\\ 70\end{array}\right]$

$=\Delta =\left|\begin{array}{ccc}6& 7& 8\\ 1& 1& 1\\ 1& 1& -1\end{array}\right|=(-6+8+7)-(8+6-7)$

$=9-7=2$

${\Delta }_1=\left|\begin{array}{ccc}100& 7& 8\\ 358& 1& 1\\ 70& 1& -1\end{array}\right|$

$=(-100+358\times 8+490)-(560+100-7\times 358)$

$=-70+15\times 358$

$=5300$

${\Delta }_2=\left|\begin{array}{ccc}6& 100& 8\\ 1& 358& 1\\ 1& 70& -1\end{array}\right|$

$=(-6\times 358+100-560)-(8\times 358+420-100)$

$=-14\times 358+660-320$

$=-4672$

${\Delta }_3=\left|\begin{array}{ccc}6& 7& 100\\ 1& 1& 358\\ 1& 1& 70\end{array}\right|$

$=(1420+100+7\times 358)-(100+6\times 358+490)$

$=288$

$x=\frac{{\Delta }_1}{\Delta }=\frac{5300}{2}=2650$

$y=\frac{{\Delta }_2}{\Delta }=\frac{-4672}{2}=2336$

$z=\frac{{\Delta }_3}{\Delta }=\frac{288}{2}=144$

Amount of each investment is Rs. $2650$, Rs. $2336$, Rs. $144$.