Question

Find the solution to the given linear system. If the system has infinite solutions, give 2 particular solutions. $$ \begin{array}{l} x_{1}+3 x_{2}+3 x_{3}=1 \\ 2 x_{1}-x_{2}+2 x_{3}=-1 \\ 4 x_{1}+5 x_{2}+8 x_{3}=2 \end{array} $$

Short answer

The system has infinite solutions. Two particular solutions are \((\frac{16}{7}, \frac{-3}{7}, 0)\) and \((\frac{-5}{7}, \frac{1}{7}, 1)\).

Step-by-step solution

Form the Augmented Matrix

Start by writing the system of equations as an augmented matrix. \[ \begin{bmatrix} 1 & 3 & 3 & | & 1 \ 2 & -1 & 2 & | & -1 \ 4 & 5 & 8 & | & 2 \end{bmatrix} \]

Row Reduction to Row Echelon Form

Perform row operations to transform the matrix into row echelon form. First, make the entry below the leading 1 in the first row zero. 1. Replace \( R_2 \) with \( R_2 - 2R_1 \): \[ \begin{bmatrix} 1 & 3 & 3 & | & 1 \ 0 & -7 & -4 & | & -3 \ 4 & 5 & 8 & | & 2 \end{bmatrix} \]2. Replace \( R_3 \) with \( R_3 - 4R_1 \): \[ \begin{bmatrix} 1 & 3 & 3 & | & 1 \ 0 & -7 & -4 & | & -3 \ 0 & -7 & -4 & | & -2 \end{bmatrix} \]

Identifying Dependent Variables and Infinite Solutions

The matrices in rows 2 and 3 are the same, illustrating a dependency and implying infinite solutions. The general form for solutions involves a parameter. Use the second row to simplify further. Replace \( R_3 \) with \( R_3 - R_2 \): \[ \begin{bmatrix} 1 & 3 & 3 & | & 1 \ 0 & -7 & -4 & | & -3 \ 0 & 0 & 0 & | & 1 \end{bmatrix} \]Now, write the solution in terms of a parameter, say \( t \):1. From row 2: \[ -7x_2 - 4x_3 = -3 \Rightarrow x_2 = \frac{-3 + 4x_3}{7} \]2. From row 1: \[ x_1 + 3\left(\frac{-3 + 4x_3}{7}\right) + 3x_3 = 1 \] Simplify to find \( x_1 \) in terms of \( x_3 \).

Determining Particular Solutions

To find particular solutions, choose specific values for the parameter \( x_3 \). 1. If \( x_3 = 0 \): - \( x_2 = \frac{-3}{7} \) - \( x_1 = 1 - 3\left(\frac{-3}{7}\right) \) calculated to be \( \frac{16}{7} \) Thus, one particular solution is \( \left( \frac{16}{7}, \frac{-3}{7}, 0 \right) \).2. If \( x_3 = 1 \): - \( x_2 = \frac{-3 + 4}{7} = \frac{1}{7} \) - \( x_1 = 1 - 3\left( \frac{1}{7} \right) - 3 \) calculated to be \( \frac{-5}{7} \) Thus, another particular solution is \( \left( \frac{-5}{7}, \frac{1}{7}, 1 \right) \).

Key concepts

Augmented Matrix

In linear algebra, an augmented matrix is a powerful tool for solving systems of linear equations. It combines the coefficients of the variables and the constants from the equations into a matrix form. For the following system of equations:

• x₁ + 3x₂ + 3x₃ = 1
• 2x₁ - x₂ + 2x₃ = -1
• 4x₁ + 5x₂ + 8x₃ = 2

we translate this into an augmented matrix:\[\begin{bmatrix}1 & 3 & 3 & \mid & 1 \2 & -1 & 2 & \mid & -1 \4 & 5 & 8 & \mid & 2\end{bmatrix}\]
This setup allows us to perform calculations more easily and efficiently. The vertical bar separates the coefficients from the constants. Transforming a system into an augmented matrix is the first step to applying techniques like row reduction.

Row Echelon Form

The process of transforming a matrix into row echelon form simplifies it, making it easier to solve the system of equations. Key steps involve creating zeros below each leading coefficient of one, forming a triangular shape.
In our augmented matrix, the first step was to clear the elements below the first pivot (1) in the first row:

• Replace row 2 with: \( R_2 - 2R_1 \)
• Replace row 3 with: \( R_3 - 4R_1 \)

These operations give us:\[\begin{bmatrix}1 & 3 & 3 & \mid & 1 \0 & -7 & -4 & \mid & -3 \0 & -7 & -4 & \mid & -2\end{bmatrix}\]This matrix highlights dependency in second and third rows, indicating potential infinite solutions. Further row operations, like simplifying row 3 to zeros, help progress towards the final solution.

Parametric Solution

When a system has infinite solutions, a parametric solution is used to express free variables in terms of one or more parameters. This involves identifying dependent equations and expressing other variables accordingly.
In our reduced system:

• From second row: \(-7x_2 - 4x_3 = -3\)

x₂ can be expressed as a function of a parameter (x₃):\[x_2 = \frac{-3 + 4x_3}{7}\]Similarly, substitute this into the first equation to find:\[x_1 + 3\left(\frac{-3 + 4x_3}{7}\right) + 3x_3 = 1\]This methodology scales down the equations to a fundamental form using a chosen parameter (here \(x_3\)), showcasing all possible solutions.

Infinite Solutions

Certain systems of linear equations don't resolve to a single solution but instead have an infinite number. This typically occurs when equations are dependent or consistent with no contradicting information.
In the example, similar rows in row echelon form indicate dependency, leading to infinite solutions. To illustrate, we derive two particular solutions using chosen values for the parameter:

• When \( x_3 = 0 \):
• When \( x_3 = 1 \):

Calculations yield particular solutions: \(\left(\frac{16}{7}, \frac{-3}{7}, 0\right)\) and \(\left(\frac{-5}{7}, \frac{1}{7}, 1\right)\).
This method highlights the flexibility of infinite solutions, offering a new perspective on analyzing systems where solutions don't fit conventional forms.