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- With these type of word problems, we need to set up the correct matrix multiplication and solve for the unknown variables.
- Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
- If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$

Let $Rs.30,000$ be divided into two parts. Assume that the amount invested in the first bond as $x$, then the amount invested in the second one is $Rs(30,000-x)$.

We are given that the first bond pays $5\%$, i,e., $0.05$ in interest and the second bond $7\%$, i.e., $0.07$.

We can represent the problem using matrix multiplication as the following $1\times 2$ matrix: $\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix}$ which is equal to the interest earned $Rs. 2000$.

Given $\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix} = 2000$:

$\begin{bmatrix}x\times (0.05)+(30000-x)\times 0.07\end{bmatrix} = 2000$

$0.05x + 2100 - 0.07x = 2000$

$2100 - 0.02x = 2000$

$2100-2000 = 0.02x$

$0.02x = 100 \rightarrow x = \frac{1}{0.02}100$

Solving for $x$, $x = 5,000$ and $30,000-x = 25,000$.

Hence the amount has to be divided into two sums of $Rs.5,000$ and $Rs.25,000$.

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