Question 1170621: Consider the following system
𝑥1 − 𝑥2 + 3𝑥3 − 𝑥4 = 0
𝑥1 + 4𝑥2 − 𝑥3 + 𝑥4 = 3
3𝑥1 + 7𝑥2 + 𝑥3 + 𝑥4 = 6
Is there any successful decomposition into endogenous and exogeneous variables? Explain.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! To determine if there's a successful decomposition into endogenous and exogenous variables, we need to analyze the system of equations and see if we can express some variables as functions of the others.
The system of equations is:
1. $x_1 - x_2 + 3x_3 - x_4 = 0$
2. $x_1 + 4x_2 - x_3 + x_4 = 3$
3. $3x_1 + 7x_2 + x_3 + x_4 = 6$
We have 3 equations and 4 variables ($x_1, x_2, x_3, x_4$). This means we have an underdetermined system, and there are infinitely many solutions (or no solutions, but we can find some).
**Endogenous and Exogenous Variables:**
* **Endogenous variables:** Variables whose values are determined within the system.
* **Exogenous variables:** Variables whose values are determined outside the system.
A successful decomposition requires that we can express some variables (endogenous) as functions of the remaining variables (exogenous).
**Analysis:**
1. **Solve for $x_1$ from equation 1:**
* $x_1 = x_2 - 3x_3 + x_4$
2. **Substitute this into equations 2 and 3:**
* Equation 2: $(x_2 - 3x_3 + x_4) + 4x_2 - x_3 + x_4 = 3$
* $5x_2 - 4x_3 + 2x_4 = 3$
* Equation 3: $3(x_2 - 3x_3 + x_4) + 7x_2 + x_3 + x_4 = 6$
* $10x_2 - 8x_3 + 4x_4 = 6$
* $5x_2 - 4x_3 + 2x_4 = 3$
3. **Notice that the resulting equations from 2 and 3 are identical.** This means we have effectively only two independent equations.
**Conclusion:**
* We can express $x_1$ as a function of $x_2, x_3, x_4$.
* We can express $5x_2 - 4x_3 + 2x_4 = 3$, which gives us one relationship between $x_2, x_3, x_4$.
Therefore, we can successfully decompose the variables.
* **Endogenous:** $x_1, x_2$ (or any two variables)
* **Exogenous:** $x_3, x_4$ (or the remaining two variables)
**Explanation:**
We can choose any two variables as exogenous. For example, if we let $x_3$ and $x_4$ be exogenous, we can solve for $x_2$ and then $x_1$.
1. $5x_2 = 4x_3 - 2x_4 + 3$
2. $x_2 = \frac{4}{5}x_3 - \frac{2}{5}x_4 + \frac{3}{5}$
3. $x_1 = (\frac{4}{5}x_3 - \frac{2}{5}x_4 + \frac{3}{5}) - 3x_3 + x_4$
4. $x_1 = -\frac{11}{5}x_3 + \frac{3}{5}x_4 + \frac{3}{5}$
Therefore, $x_1$ and $x_2$ are successfully expressed as functions of $x_3$ and $x_4$.
**Final Answer:** Yes, there is a successful decomposition into endogenous and exogenous variables. For example, we can choose $x_1$ and $x_2$ as endogenous and $x_3$ and $x_4$ as exogenous.
|
|
|
