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Let $x$ be the number of ₹ 1,000 vouchers and $y$ be the number of ₹ 500 vouchers.
Based on the problem statement: $$x + y = 60$$ $$x + 3y = 100$$
The system can be written as $AX = B$, where: $$A = \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix}, X = \begin{pmatrix} x \\ y \end{pmatrix}, B = \begin{pmatrix} 60 \\ 100 \end{pmatrix}$$
Calculate determinant $|A| = (1)(3) - (1)(1) = 2$. Since $|A| \neq 0$, $A^{-1}$ exists. $$A^{-1} = \frac{1}{2} \begin{pmatrix} 3 & -1 \\ -1 & 1 \end{pmatrix}$$ $$X = A^{-1}B = \frac{1}{2} \begin{pmatrix} 3 & -1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 60 \\ 100 \end{pmatrix}$$ $$X = \frac{1}{2} \begin{pmatrix} 180 - 100 \\ -60 + 100 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 80 \\ 40 \end{pmatrix} = \begin{pmatrix} 40 \\ 20 \end{pmatrix}$$
Final Answer: Number of ₹ 1,000 vouchers = 40, Number of ₹ 500 vouchers = 20
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