Math Problem Statement

The following system of equations has: x1 + x3 = 0, x2 - x3 = 0, x1 + x2 = 0. Does the system have: A. One solution, B. More than one solution, C. No solution?

Solution

The system of equations given is:

x_1 + x_3 &= 0 \\ x_2 - x_3 &= 0 \\ x_1 + x_2 &= 0 \end{aligned}$$ We need to determine if the system has: - A: One solution. - B: More than one solution. - C: No solution. ### Step 1: Express variables in terms of one another From equation $$x_1 + x_3 = 0$$, we have: $$x_1 = -x_3$$ From equation $$x_2 - x_3 = 0$$, we have: $$x_2 = x_3$$ ### Step 2: Substitute into the third equation Substituting $$x_1 = -x_3$$ and $$x_2 = x_3$$ into the third equation $$x_1 + x_2 = 0$$: $$-x_3 + x_3 = 0$$ This holds true, meaning that all three equations are consistent. ### Step 3: Conclusion Since there are three variables ($$x_1, x_2, x_3$$) and only two independent equations (the third equation doesn't add new information), the system is underdetermined and has **infinitely many solutions**. The general solution will depend on the free variable $$x_3$$. Thus, the correct answer is: - **B: More than one solution.** Would you like further details on this process or clarification on any step? --- Here are 5 related questions to explore further: 1. How do you solve systems of equations using matrix methods? 2. What does it mean for a system to be consistent or inconsistent? 3. How can you determine the number of solutions to a system of linear equations? 4. What role does the rank of a matrix play in solving linear systems? 5. How does Gaussian elimination help in solving systems of equations? **Tip**: For underdetermined systems, always check for free variables by comparing the number of equations to the number of unknowns.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Underdetermined Systems
Infinitely Many Solutions

Formulas

Linear system: ax + by + cz = d

Theorems

The rank-nullity theorem
Consistent and Inconsistent systems

Suitable Grade Level

Undergraduate/Advanced High School (Grades 11-12)

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