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Question
A total amount of ₹7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and \[8\frac{1}{2}\] % respectively. The total annual interest from these three accounts is ₹550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.
Solution
Let the amount deposited in each of the three accounts be ₹ x, ₹ x and ₹ y respectively.
Since, the total amount deposited is ₹7,000.
∴ x + x + y = 7000
⇒ 2x + y = 7000 ....(1)
Total annual Interest is ₹550.
∴ \[\frac{5}{100}x + \frac{8}{100}x + \frac{17}{200}y = 550\]
The above system of equations can be written in matrix form AX = B as \[\begin{bmatrix}2 & 1 \\ 26 & 17\end{bmatrix}\binom{x}{y} = \binom{7000}{110000}\]
\[\text{ where,} A = \begin{bmatrix}2 & 1 \\ 26 & 17\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{7000}{110000}\]
Now,
\[\left| A \right| = \begin{vmatrix}2 & 1 \\ 26 & 17\end{vmatrix}\]
\[ = 34 - 26\]
\[ = 8\]
\[\text{ Let }C_{ij}\text{ be the cofactors of elements }a_{ij}\text{ in }A = \left[ a_{ij} \right] .\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} 17 = 17, C_{12} = \left( - 1 \right)^{1 + 2} 26 = - 26\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} 1 = - 1 , C_{22} = \left( - 1 \right)^{2 + 2} 2 = 2\]
\[adj A = \begin{bmatrix}17 & - 26 \\ - 1 & 2\end{bmatrix}^T \]
\[ = \begin{bmatrix}17 & - 1 \\ - 26 & 2\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{8}\begin{bmatrix}17 & - 1 \\ - 26 & 2\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{8}\begin{bmatrix}17 & - 1 \\ - 26 & 2\end{bmatrix}\binom{7000}{110000}\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{8}\binom{119000 - 110000}{ - 182000 + 220000}\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{8}\binom{9000}{38000}\]
\[ \Rightarrow x = \frac{9000}{8}\text{ and }y = \frac{38000}{8}\]
\[ \therefore x = 1125\text{ and }y = 4750 .\]
Hence, the amount deposited in each of the three accounts is ₹1125, ₹1125 and ₹4750.
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