Problem 28 - Differential Equations and Linear Algebra

Question

Show that the system $$ \begin{aligned} x_{1}+x_{2}+x_{3} &=y_{1} \\ 2 x_{1}+3 x_{2}+x_{3} &=y_{2} \\ 3 x_{1}+5 x_{2}+x_{3} &=y_{3} \end{aligned} $$ has an infinite number of solutions provided that \(\left(y_{1}, y_{2}, y_{3}\right)\) lies on the plane with equation \(y_{1}-2 y_{2}+\) \(y_{3}=0\)

Step-by-Step Solution

Answer

The row echelon form of the augmented matrix for the given linear system is \[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & y_1 \\ 0 & 1 & -1 & y_2-y_1 \\ 0 & 0 & 0 & y_1-2y_2+y_3 \end{array} \right] \] Under the condition \(y_1-2y_2+y_3=0\), the last row becomes redundant, and we can express the solutions in terms of an arbitrary constant \(t\): \[ \begin{aligned} x_1 &= y_1 - y_2 + y_1 - t, \\ x_2 &= y_2 - y_1 + t, \\ x_3 &= t, \end{aligned} \] where \(t\) can be any real number. Since there is a free variable, the system has an infinite number of solutions.

Step 1: Write the augmented matrix

Begin by representing the given linear system of equations in an augmented matrix: \[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & y_1 \\ 2 & 3 & 1 & y_2 \\ 3 & 5 & 1 & y_3 \end{array} \right] \]

Step 2: Perform Gaussian elimination

Perform Gaussian elimination to bring the augmented matrix to row echelon form. First, subtract the first row from the second row, and then subtract 3 times the first row from the third row: \[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & y_1 \\ 0 & 1 & -1 & y_2-y_1 \\ 0 & 2 & -2 & y_3-3y_1 \end{array} \right] \] Next, subtract 2 times the second row from the third row: \[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & y_1 \\ 0 & 1 & -1 & y_2-y_1 \\ 0 & 0 & 0 & y_3-3y_1-2(y_2-y_1) \end{array} \right] \]

Step 3: Analyze the row echelon form

In row echelon form, we can see that the last row indicates that the given condition on the \(y_i\)s is met: \(y_3-3y_1-2(y_2-y_1) = y_1-2y_2+y_3 = 0\). Under this condition, we can ignore the last row, as it does not provide any information about the solutions for \(x_1, x_2, x_3\). Rewriting the second and first rows as equations, we get \[ \begin{aligned} x_2 - x_3 &= y_2 - y_1, \\ x_1 + x_2 + x_3 &= y_1. \end{aligned} \]

Step 4: Express the solutions in terms of arbitrary constants

To show that the system has an infinite number of solutions, we can rewrite the solutions in terms of arbitrary constants. Let \(x_3 = t\). Then, \(x_2 = y_2 - y_1 + t\) and \(x_1 = y_1 - (y_2 - y_1 + t) - t = y_1 - y_2 + y_1 - t\). The solutions to the system are given by \[ \begin{aligned} x_1 &= y_1 - y_2 + y_1 - t, \\ x_2 &= y_2 - y_1 + t, \\ x_3 &= t, \end{aligned} \] where \(t\) can be any real number. Since there is a free variable, there are an infinite number of solutions to the system, which depend on the values of \((y_1, y_2, y_3)\) that satisfy the given plane equation.

Key Concepts

Gaussian Elimination

Gaussian Elimination is a systematic method used in linear algebra to solve systems of linear equations. It transforms a given system of equations into a simpler form, known as row echelon form, making it easier to find solutions.
This method involves using a sequence of operations:

• Swapping rows.
• Multiplying a row by a nonzero constant.
• Adding or subtracting a multiple of one row from another.

In the given problem, the augmented matrix represents a system of three linear equations. By applying Gaussian Elimination, we simplify this matrix step by step until it's in row echelon form.
The process helps to systematically isolate one variable at a time, simplifying the equations to ultimately solve for all variables. The last step typically reveals whether the system has no solution, a single solution, or infinitely many solutions.

System of Equations

A System of Equations consists of multiple equations that are solved simultaneously. Each equation represents a hyperplane in the n-dimensional space, where

• The intersection of these hyperplanes, if it exists, corresponds to the solution(s) of the system.
• If the equations describe parallel hyperplanes, the system might have no solution.
• If the hyperplanes intersect at a line or a plane, they represent infinite solutions.

In the given exercise, the system consists of three equations with three variables: \(\begin{aligned} x_{1}+x_{2}+x_{3} &=y_{1} \ 2 x_{1}+3 x_{2}+x_{3} &=y_{2} \ 3 x_{1}+5 x_{2}+x_{3} &=y_{3} \end{aligned}\)
As the solutions progress, if the last row is entirely zeros as in the row-echeloned form, it hints at the possibility of infinite solutions, given certain conditions are met.

Infinite Solutions

An Infinite Solution Set in a system of equations means that there are countless solutions meeting the given criteria. This usually happens when there is a free variable, allowing flexibility in assigning values.
For a system to have infinite solutions:

• The rows of the matrix in row-echelon form must indicate a dependency without contradiction.
• Any additional conditions must be met—for example, in this problem: \( y_1-2y_2+y_3=0 \).

This condition results in one equation being redundant, thus creating a dependency that allows an arbitrary choice for at least one variable.
The exercise demonstrates this by choosing free variable \( x_3 = t \), resulting in expressions for \( x_1 \) and \( x_2 \) in terms of \( t \). With \( t \) freely chosen, each choice corresponds to a different solution to the system, thus providing an infinite number of solutions. Each solution corresponds to a different position along a line or plane within the solution space.