Question

Find the solution to the given linear system. If the system has infinite solutions, give 2 particular solutions. $$ \begin{aligned} 2 x_{1}+4 x_{2}+6 x_{3} &=2 \\ 1 x_{1}+2 x_{2}+3 x_{3} &=1 \\ -3 x_{1}-6 x_{2}-9 x_{3} &=-3 \end{aligned} $$

Short answer

The system has infinite solutions; two particular solutions are (1, 0, 0) and (-1, 1, 0).

Step-by-step solution

Write the System in Matrix Form

First, write the system of equations as an augmented matrix: \[\begin{bmatrix}2 & 4 & 6 & | & 2 \1 & 2 & 3 & | & 1 \-3 & -6 & -9 & | & -3\end{bmatrix}\]

Perform Row Operations

Use row operations to simplify the matrix. First, divide the second row by 1 (it's already simplified): \[R_2 \leftarrow R_2\]Next, add 3 times the second row to the third row to eliminate the first column's terms:\[R_3 \leftarrow R_3 + 3 \times R_2\]This gives:\[\begin{bmatrix}2 & 4 & 6 & | & 2 \1 & 2 & 3 & | & 1 \0 & 0 & 0 & | & 0\end{bmatrix}\]

Further Simplify the Matrix

Subtract twice the second row from the first row:\[R_1 \leftarrow R_1 - 2 \times R_2\]This reduces the matrix to:\[\begin{bmatrix}0 & 0 & 0 & | & 0 \1 & 2 & 3 & | & 1 \0 & 0 & 0 & | & 0\end{bmatrix}\]

Interpret the Simplified Matrix

The second row gives the equation:\[x_1 + 2x_2 + 3x_3 = 1\]Since the other rows are zero, this is the only condition for the solutions, indicating a system with infinite solutions.

Express Solutions in Terms of Parameters

Let \(x_2 = s\) and \(x_3 = t\), where \(s\) and \(t\) are parameters. Then express \(x_1\) as:\[x_1 = 1 - 2s - 3t\]

Identify Two Particular Solutions

Choose specific values for \(s\) and \(t\) to find particular solutions. For example, let \(s = 0\), \(t = 0\):\[x_1 = 1, x_2 = 0, x_3 = 0\] so one solution is \((1, 0, 0)\).Let \(s = 1\), \(t = 0\):\[x_1 = -1, x_2 = 1, x_3 = 0\] so another solution is \((-1, 1, 0)\).

Key concepts

Matrix Row Operations

Matrix row operations are essential for solving systems of linear equations. They allow us to simplify matrices and find solutions efficiently.

There are three main types of row operations:

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

These operations help create zeros in strategic places, which can make the system easier to solve.

It’s often used to transform the matrix into Row Echelon Form (REF) or Reduced Row Echelon Form (RREF). In this example, we start by organizing the system as an augmented matrix and then proceeded by adding multiples of rows to simplify the rows until the system is easier to interpret.

In the example, we first restructured the third row to be all zeros by adding a multiple of another row. By step-by-step iteration, we made the matrix simpler.

Infinite Solutions

Infinite solutions in a linear system occur when there is not a unique solution. Instead, there is a family of solutions that satisfy the given equations. This usually happens when at least one equation in the system can be expressed as a linear combination of the others.

In our problem, we simplified the matrix such that two rows turned into zero rows without affecting the consistency of the system. This led to the realization that there are infinitely many solutions. Essentially, instead of pinpointing a single set of values for the variables, the solution set forms a line or a plane in the coordinate space, which contains all the points that solve the system.

Systems with infinite solutions are commonly characterized by having free variables. They offer a range of solutions based on different variable assignments, allowing for dynamic understanding of the system's constraints.

Parameterization of Solutions

Parameterization involves expressing solutions of a system in terms of one or more free parameters. This is a technique often used when a system has infinitely many solutions.

In our problem, after simplifying the matrix, we found that the equation was:
x_1 + 2x_2 + 3x_3 = 1.

We selected free parameters, let's say, letting \(x_2 = s\) and \(x_3 = t\). This allows us to express \(x_1\) as a function of these parameters:
\(x_1 = 1 - 2s - 3t\).

By doing this, you transform the originally infinite set of solutions into an easy-to-handle form. Any specific value of \(s\) and \(t\) gives a particular solution. This method is very useful in understanding how the solutions relate to the parameters, enabling you to easily find different instances of the solution.

Augmented Matrix

An augmented matrix is used to represent a system of linear equations. This matrix includes the coefficients of the variables and the constants from the equations in a single matrix format.

In this specific example, our system is represented by the augmented matrix: \[\begin{bmatrix}2 & 4 & 6 & | & 2 \1 & 2 & 3 & | & 1 \-3 & -6 & -9 & | & -3\end{bmatrix}.\]

The vertical line in the matrix separates the coefficients matrix from the constants column. An augmented matrix is essential because it consolidates all the elements of the linear equations together, making it easier to manipulate using row operations.

This form allows us to apply mathematical procedures systemically until the system is simplified or solved, facilitating both manual and computational methods to derive solutions. It’s a fundamental and strategic way of organizing data to depict the structure of the system uniquely.