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, $x + y + z = 5000$ $\quad...\left(1\right)$
$\frac{6x}{100}+\frac{7y}{100}+\frac{8z}{100}=358$
$\Rightarrow \quad6x + 7y + 8z = 35800$ $\quad...\left(2\right)$
And, $\frac{6x}{100}+\frac{7y}{100}-\frac{8z}{100}=70$
$\Rightarrow \quad6x + 7y - 8z = 7000$ $\quad...\left(3\right)$
The equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$ can be written in the form $AX = B$ as
$\left[\begin{matrix}1&1&1\\ 6&7&8\\ 6&7&-8\end{matrix}\right]\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\left[\begin{matrix}5000\\ 35800\\ 7000\end{matrix}\right]$

Applying $R_{2}\rightarrow R_{2}-6R_{1}, R_{3}\rightarrow R_{3}-6R_{1}$, we get
$\left[\begin{matrix}1&1&1\\ 0&1&2\\ 0&1&-14\end{matrix}\right]\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\left[\begin{matrix}5000\\ 5800\\ -23000\end{matrix}\right]$

Applying $R_{3}\rightarrow R_{3}- R_{2}$, we get
$\left[\begin{matrix}1&1&1\\ 0&1&2\\ 0&0&-16\end{matrix}\right]\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\left[\begin{matrix}5000\\ 5800\\ -22800\end{matrix}\right]$

Now as A is reduced to an upper triangular matrix, we rewrite the equation in the original form as
$x + y + z = 5000\quad\ldots\left(i\right)$
$y + 2z = 5800\quad\ldots\left(ii\right)$
$- 1 6 z = - 28800\quad\qquad...\left(iii\right)$
From $\left(iii\right)$, $z = 1800$, from $\left(ii\right)$, $y = 2200$
From $\left(i\right)$, $x = 1000$
$\therefore \quad$ The amount of first, second and third investments is $₹ 1000$, $₹ 2200$ and $₹ 1800$ respectively.