Question

An amount of ₹ 5000 is put into three investments of 6%, 7% and 8% per annum respectively. The total annual income from these investments is ₹ 358. If the total annual income from first two investments is ₹ 70 more than the income from the third, find the amount of each investment by the matrix method.

• A. $₹500$, $₹2200$, $₹1800$
• B. $₹1000$, $₹2200$, $₹1800$
• C. $₹1200$, $₹1400$, $₹2000$
• D. none of these

Answer 1

Answer: (B)

Let the first, second and third investments be $₹x$, $₹y$ and $₹z$ respectively.
Then, $x+y+z=5000$ $...\left(1\right)$
$\frac{6x}{100}+\frac{7y}{100}+\frac{8z}{100}=358$
$\Rightarrow 6x+7y+8z=35800$ $...\left(2\right)$
And, $\frac{6x}{100}+\frac{7y}{100}-\frac{8z}{100}=70$
$\Rightarrow 6x+7y-8z=7000$ $...\left(3\right)$
The equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$ can be written in the form $AX=B$ as
$\left[\begin{array}{ccc}1& 1& 1\\ 6& 7& 8\\ 6& 7& -8\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5000\\ 35800\\ 7000\end{array}\right]$

Applying $R_2\to R_2-6R_1,R_3\to R_3-6R_1$, we get
$\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 1& -14\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5000\\ 5800\\ -23000\end{array}\right]$

Applying $R_3\to R_3-R_2$, we get
$\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 0& -16\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5000\\ 5800\\ -22800\end{array}\right]$

Now as A is reduced to an upper triangular matrix, we rewrite the equation in the original form as
$x+y+z=5000\dots \left(i\right)$
$y+2z=5800\dots \left(ii\right)$
$-16z=-28800...\left(iii\right)$
From $\left(iii\right)$, $z=1800$, from $\left(ii\right)$, $y=2200$
From $\left(i\right)$, $x=1000$
$\therefore$ The amount of first, second and third investments is $₹1000$, $₹2200$ and $₹1800$ respectively.