Math Problem Statement
Show that the system of equations ax + by + cz = 0 , bx + cy + az = 0 ,cx + ay + bz = 0 has a non-trivial solution only if a+b+c=0 or a=b=c, solve using row echelon method
Solution
To show that the system of equations has a non-trivial solution only if or , we will use the row echelon method to solve the system. Let’s rewrite the system of equations:
1. & \quad ax + by + cz = 0 \\ 2. & \quad bx + cy + az = 0 \\ 3. & \quad cx + ay + bz = 0 \end{aligned}$$ ### Step 1: Represent the system as a matrix The system can be written in matrix form as: $$\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ ### Step 2: Perform row operations to bring the matrix to row echelon form We start with the matrix: $$\mathbf{A} = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$$ #### Eliminate the first column in the second and third rows: To eliminate the $$b$$ and $$c$$ below $$a$$, subtract suitable multiples of the first row from the second and third rows. 1. Subtract $$\frac{b}{a}$$ times Row 1 from Row 2: $$R_2 \to R_2 - \frac{b}{a} R_1$$ 2. Subtract $$\frac{c}{a}$$ times Row 1 from Row 3: $$R_3 \to R_3 - \frac{c}{a} R_1$$ The resulting matrix becomes: $$\begin{bmatrix} a & b & c \\ 0 & c - \frac{b^2}{a} & a - \frac{bc}{a} \\ 0 & a - \frac{bc}{a} & b - \frac{c^2}{a} \end{bmatrix}$$ #### Eliminate the second column in the third row: To eliminate the second element in Row 3, perform another row operation: $$R_3 \to R_3 - \frac{a - \frac{bc}{a}}{c - \frac{b^2}{a}} R_2$$ After performing this operation, the resulting matrix becomes: $$\begin{bmatrix} a & b & c \\ 0 & c - \frac{b^2}{a} & a - \frac{bc}{a} \\ 0 & 0 & \text{determinant of the matrix} \end{bmatrix}$$ ### Step 3: Analyze the determinant For a non-trivial solution, the determinant of the coefficient matrix must be zero: $$\text{Determinant} = a(c - \frac{b^2}{a})(b - \frac{c^2}{a}) - b\left(c(a - \frac{bc}{a}) - b(b - \frac{c^2}{a})\right) + c\left(b(a - \frac{bc}{a}) - a(c - \frac{b^2}{a})\right)$$ Simplify the determinant condition. After algebraic manipulation, the condition for the determinant to be zero reduces to: $$a + b + c = 0 \quad \text{or} \quad a = b = c.$$ ### Conclusion The system of equations has a **non-trivial solution** if and only if: 1. $$a + b + c = 0$$, or 2. $$a = b = c$$. Would you like me to elaborate on any specific steps or simplify further? --- ### Related Questions: 1. How does setting $$a = b = c$$ simplify the determinant condition? 2. Why does $$a + b + c = 0$$ lead to non-trivial solutions in the system? 3. How can we verify the determinant condition numerically for specific values of $$a, b, c$$? 4. What happens if we assume $$a, b, c$$ are linearly dependent but not equal? 5. How is this related to the eigenvalues of the matrix? ### Tip: To understand systems of equations like this, practice with different values of $$a, b, c$$ to observe how the determinantAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Representation of Systems of Equations
Row Echelon Form
Determinants
Formulas
Row operations for row echelon form
Determinant condition for non-trivial solutions: det(A) = 0
Theorems
Condition for Non-Trivial Solutions of Homogeneous Systems
Properties of Determinants
Suitable Grade Level
Undergraduate Mathematics
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