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# 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.

Last updated date: 17th Sep 2024
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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.

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}