Math Problem Statement
find the no. of distinct pairs of m,n such that the following equations have no unique solution 8x + m(y) = 13 n(x) + 5y = 18
Solution
We are tasked with finding the number of distinct pairs such that the system of linear equations:
has no unique solution. For a system of two linear equations to have no unique solution, the system must either be inconsistent (no solution) or have infinitely many solutions (dependent equations). This occurs when the determinant of the coefficient matrix is zero.
The coefficient matrix of the system is:
8 & m \\ n & 5 \end{pmatrix}$$ The determinant of this matrix is given by: $$\text{Determinant} = 8 \times 5 - m \times n = 40 - mn$$ For the system to have no unique solution, we require the determinant to be zero: $$40 - mn = 0$$ This simplifies to: $$mn = 40$$ Thus, we need to find all distinct pairs $$(m, n)$$ such that their product equals 40. ### Finding pairs $$(m, n)$$ We now list all factorizations of 40: $$40 = 1 \times 40, \quad 2 \times 20, \quad 4 \times 10, \quad 5 \times 8$$ In addition to positive factors, we also have negative factor pairs: $$40 = (-1) \times (-40), \quad (-2) \times (-20), \quad (-4) \times (-10), \quad (-5) \times (-8)$$ Thus, the distinct pairs $$(m, n)$$ are: $$(1, 40), (2, 20), (4, 10), (5, 8), (-1, -40), (-2, -20), (-4, -10), (-5, -8)$$ Additionally, we must account for the fact that the order of $$(m, n)$$ matters. Thus, we must also include the reverse pairs, like $$(40, 1)$$, etc. The complete set of pairs is: $$(1, 40), (40, 1), (2, 20), (20, 2), (4, 10), (10, 4), (5, 8), (8, 5), (-1, -40), (-40, -1), (-2, -20), (-20, -2), (-4, -10), (-10, -4), (-5, -8), (-8, -5)$$ There are a total of 16 distinct pairs $$(m, n)$$. ### Conclusion: The number of distinct pairs $$(m, n)$$ such that the system has no unique solution is **16**. Would you like further details or have any questions? --- Here are some related questions to explore: 1. How do you find the determinant of a 3x3 matrix? 2. What is the condition for a system of equations to have infinitely many solutions? 3. Can a system of linear equations with a non-zero determinant ever have no solution? 4. How does Cramer's Rule relate to systems with unique solutions? 5. What are the properties of matrices that determine the number of solutions of a system? **Tip:** The determinant is a key concept in linear algebra for solving systems of equations. A non-zero determinant implies a unique solution, while a zero determinant indicates either no solution or infinitely many solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Determinants
Systems of Linear Equations
Formulas
Determinant of a 2x2 matrix: Det(A) = ad - bc
Condition for no unique solution: determinant = 0
Theorems
Condition for systems of linear equations to have no unique solution (determinant = 0)
Suitable Grade Level
Grades 10-12
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