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.By row echelon method

Solution

We will analyze the system of equations

ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0ax + by + cz = 0, \quad bx + cy + az = 0, \quad cx + ay + bz = 0

using the row echelon method to show that the system has a non-trivial solution only if a+b+c=0a + b + c = 0 or a=b=ca = b = c.

Step 1: Representing the system as a matrix equation

The system of equations can be written in matrix form as:

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}.$$ The coefficient matrix is: $$A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}.$$ The determinant of $$A$$ will determine whether the system has a non-trivial solution. If $$\det(A) = 0$$, the system has a non-trivial solution. ### Step 2: Row reduction of $$A$$ to row echelon form Start with $$A$$: $$A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}.$$ #### Subtract multiples of Row 1 from Rows 2 and 3: - Replace Row 2 ($$R_2$$) with $$R_2 - \frac{b}{a} R_1$$ (assuming $$a \neq 0$$). - Replace Row 3 ($$R_3$$) with $$R_3 - \frac{c}{a} R_1$$. The matrix becomes: $$\begin{bmatrix} a & b & c \\ 0 & \frac{ac - b^2}{a} & \frac{ab - bc}{a} \\ 0 & \frac{ab - ac}{a} & \frac{ab - c^2}{a} \end{bmatrix}.$$ #### Simplify second and third rows: To simplify the analysis, denote the second row's diagonal element as $$D_2 = \frac{ac - b^2}{a}$$, and analyze the determinant. ### Step 3: Compute the determinant of $$A$$ The determinant of $$A$$ is: $$\det(A) = a \det\begin{bmatrix} c & a \\ a & b \end{bmatrix} - b \det\begin{bmatrix} b & a \\ c & b \end{bmatrix} + c \det\begin{bmatrix} b & c \\ c & a \end{bmatrix}.$$ Perform the calculations step by step: 1. $$\det\begin{bmatrix} c & a \\ a & b \end{bmatrix} = cb - a^2$$, 2. $$\det\begin{bmatrix} b & a \\ c & b \end{bmatrix} = b^2 - ac$$, 3. $$\det\begin{bmatrix} b & c \\ c & a \end{bmatrix} = ab - c^2$$. Substitute into the determinant: $$\det(A) = a(cb - a^2) - b(b^2 - ac) + c(ab - c^2).$$ Simplify: $$\det(A) = acb - a^3 - b^3 + abc + abc - c^3.$$ Combine terms: $$\det(A) = a^3 + b^3 + c^3 - 3abc.$$ ### Step 4: Factorization of the determinant The determinant can be factorized as: $$\det(A) = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).$$ ### Step 5: Conditions for $$\det(A) = 0$$ For the system to have a non-trivial solution, $$\det(A) = 0$$. This happens if: 1. $$a + b + c = 0$$, or 2. $$a^2 + b^2 + c^2 - ab - bc - ca = 0$$. #### Analyzing the second condition: The expression $$a^2 + b^2 + c^2 - ab - bc - ca$$ can be rewritten as: $$\frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2].$$ This is zero if and only if $$a = b = c$$. ### Final Conclusion: The system of equations has a non-trivial solution only if: 1. $$a + b + c = 0$$, or 2. $$a = b = c$$. --- Would you like to see more details or work through an example with specific numbers? Here are five related questions you might explore: 1. How do we check for other non-trivial solutions using eigenvalues? 2. What happens if $$a = b = c = 0$$ in this system? 3. Can we find the general solution when $$a + b + c = 0$$? 4. How does the symmetry of the coefficient matrix affect its determinant? 5. How can we verify these results geometrically? **Tip:** When analyzing systems of equations, symmetry in the coefficient matrix can often provide clues about simplifications and

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
System of Linear Equations
Row Echelon Form
Determinants

Formulas

Matrix determinant formula: det(A) = Σ (a(i,j) * C(i,j))
Factorization of determinant: a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
Simplified determinant condition: det(A) = 0

Theorems

Row Reduction Theorem
Determinant Factorization
Symmetry Property of Determinants

Suitable Grade Level

Undergraduate Level