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Question
A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2800 as interest. However, if trust had interchanged money in bonds, they would have got ₹ 100 less as interes. Using matrix method, find the amount invested by the trust.
Solution
Let Rs x be invested in the first bond and Rs y be invested in the second bond.
Let A be the investment matrix and B be the interest per rupee matrix. Then,
A =[x y ]` and `B = [[10/100],[12/100]]`
Total annual interest = AB= [ x y ] `[[10/100],[12/100]]` = `(10x)/100+(12y)/100`
\[ \therefore \frac{10x}{100} + \frac{12y}{100} = 2800\]
\[ \Rightarrow 10x + 12y = 280000 . . . . . \left( 1 \right)\]
If the rates of interest had been interchanged, then the total interest earned is Rs 100 less than the previous interest
\[\therefore \frac{12x}{100} + \frac{10y}{100} = 2700\]
\[ \Rightarrow 12x + 10y = 270000 . . . . . \left( 2 \right)\]
The system of equations (1) and (2) can be expressed as
PX = Q, where
`P = [[ 10 12],[12 10]]`, ` X =[[x ],[y]]`, `Q = [[28000],[27000]]`
\[\left| P \right| = \begin{vmatrix}10 & 12 \\ 12 & 10\end{vmatrix} = 100 - 144 = - 44 \neq 0\]
Thus, P is invertible.
\[\therefore X = P^{- 1} Q\]
\[ \Rightarrow X = \frac{adj P}{\left| P \right|}Q\]
`⇒ [[x ],[y]]` = `1/((-44))` `[[ 10 -12] ,[ -12 10]]^T` `[[280000],[270000]]`
`⇒ [[x ],[y]]` = `1/((-44))` `[[ 10 -12] ,[ -12 10]]` `[[280000],[270000]]`
`⇒ [[x ],[y]]=[[(2800000 - 3240000)/(- 44)],[(- 3360000 + 2700000)/(- 44)]]` = `[[10000],[15000]]`
\[ \Rightarrow x = \text{10000 and y }= 15000\]
Therefore, Rs 10,000 be invested in the first bond and Rs 15,000 be invested in the second bond.
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