Question

A trust caring for handicapped children gets Rs 30000 every month from its donors. The trust spends half of the funds received for medical & educational care of the children & for that it charges 2% of the spent amount from them, & deposited the balance amount in a private bank to get the money multiplied so that the trust goes on functioning regularly. What percent of interest should the trust get from the bank so as to get a total of Rs. 1800 every month? Use the matrix method to find the rate of interest.

Answer 1

Hint: Let us assume that the earning of trust from all the charges is x and earning of trust from the interest obtained from the bank is y. It is given that the total earning of trust is Rs. 1800. So we get equation as: $x+y=1800$
Also, it is said that the trust receives Rs 30000 every month out of which it spends half on medical & educational care of the children, i.e. Rs 15000 and deposits rest to the bank. It also charges 2% of this amount to the bank. So, the earning of trust from these charges is: 2% of Rs. 15000, i.e. Rs. 300. So, we have another equation: $x=300$
So, for a system of linear equations in two variables:
$\begin{align}
  & ax+by=p \\
 & cx+dy=q \\
\end{align}$
We can solve by using matrix method as:
$\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   p \\
   q \\
\end{matrix} \right]$
Solve the matrix for the given equations to get the value of y. Now, assume that the interest rate is r%. So, by using the formula: $\text{interest}=rate\times amount$ , where y is the interest and Rs. 15000 is the amount, get the value of interest rate.

Complete step by step answer:
As we have a system of linear equations in two variables:
$\begin{align}
  & x+y=1800......(1) \\
 & x=300......(2) \\
\end{align}$
So, by using matrix method, we can write:
$\left[ \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   1800 \\
   300 \\
\end{matrix} \right]......(3)$
Now, by separating the variables on one side, we can write:
\[\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]={{\left[ \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right]}^{-1}}\left[ \begin{matrix}
   1800 \\
   300 \\
\end{matrix} \right]......(4)\]
For a matrix: $A=\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]$ , we have \[{{A}^{-1}}=\dfrac{adjA}{\left| A \right|}\], where $adjA={{\left[ {{C}_{i}}_{j} \right]}^{T}}$ and ${{C}_{i}}_{j}={{\left( -1 \right)}^{i+j}}{{M}_{i}}_{j}$
So, for matrix: $A=\left[ \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right]$
$\begin{align}
  & \left| A \right|=-1 \\
 & {{M}_{i}}_{j}=\left[ \begin{matrix}
   0 & 1 \\
   1 & 1 \\
\end{matrix} \right] \\
 & {{C}_{i}}_{j}={{\left( -1 \right)}^{i+j}}\left[ \begin{matrix}
   0 & 1 \\
   1 & 1 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   0 & -1 \\
   -1 & 1 \\
\end{matrix} \right] \\
 & adjA={{\left[ \begin{matrix}
   0 & -1 \\
   -1 & 1 \\
\end{matrix} \right]}^{T}} \\
 & =\left[ \begin{matrix}
   0 & -1 \\
   -1 & 1 \\
\end{matrix} \right]
\end{align}$
Therefore,
\[\begin{align}
  & {{A}^{-1}}=\dfrac{1}{-1}\left[ \begin{matrix}
   0 & -1 \\
   -1 & 1 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   0 & 1 \\
   1 & -1 \\
\end{matrix} \right]
\end{align}\]
So, we can write equation (4) as:
\[\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 1 \\
   1 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1800 \\
   300 \\
\end{matrix} \right]......(5)\]

Now, solve the matrix, to get the value of y:
\[\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   \left( 0\times 1800 \right)+\left( 1\times 300 \right) \\
   \left( 1\times 1800 \right)+\left( -1\times 300 \right) \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   300 \\
   1500 \\
\end{matrix} \right] \\
\end{align}\]
Therefore, y = Rs. 1500
Now, by using the formula: $\text{interest}=rate\times amount$, we get the interest rate r as:
$\begin{align}
  & \Rightarrow 1500=\dfrac{r}{100}\times 15000 \\
 & \Rightarrow r=10\% \\
\end{align}$

Hence, the interest rate is 10%

Note: There is an alternate method for solving the inverse of a matrix, i.e.
For a matrix: $A=\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]$ , we have \[{{A}^{-1}}=\dfrac{1}{ad-bc}\left[ \begin{matrix}
   d & -c \\
   -b & a \\
\end{matrix} \right]\]
So, we can write equation (4) as:
\[\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\dfrac{1}{\left( 1\times 0 \right)-\left( 1\times 1 \right)}\left[ \begin{matrix}
   0 & -1 \\
   -1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1800 \\
   300 \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\dfrac{1}{-1}\left[ \begin{matrix}
   0 & -1 \\
   -1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1800 \\
   300 \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 1 \\
   1 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1800 \\
   300 \\
\end{matrix} \right] \\
\end{align}\]