Chapter 1: Problem 13
Find the solution to the given linear system. If the system has infinite solutions, give 2 particular solutions. $$ \begin{array}{l} x_{1}+2 x_{2}+2 x_{3}=1 \\ 2 x_{1}+x_{2}+3 x_{3}=1 \\ 3 x_{1}+3 x_{2}+5 x_{3}=2 \end{array} $$
Short Answer
Expert verified
The system has infinite solutions. Two particular solutions are \((\frac{1}{3}, \frac{1}{3}, 0)\) and \((-1, 0, 1)\).
Step by step solution
01
Write the System as a Matrix Equation
First, express the system as a matrix equation. The matrix form for this linear system is: \[ A \vec{x} = \vec{b} \]where \( A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 3 \ 3 & 3 & 5 \end{bmatrix} \), \( \vec{x} = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} \), and \( \vec{b} = \begin{bmatrix} 1 \ 1 \ 2 \end{bmatrix} \).
02
Perform Row Reduction
Use Gaussian elimination to row-reduce the matrix. Start with the augmented matrix:\[ \begin{bmatrix} 1 & 2 & 2 & | & 1 \ 2 & 1 & 3 & | & 1 \ 3 & 3 & 5 & | & 2 \end{bmatrix} \]Subtract 2 times the first row from the second row:\[ \begin{bmatrix} 1 & 2 & 2 & | & 1 \ 0 & -3 & -1 & | & -1 \ 3 & 3 & 5 & | & 2 \end{bmatrix} \]Subtract 3 times the first row from the third row:\[ \begin{bmatrix} 1 & 2 & 2 & | & 1 \ 0 & -3 & -1 & | & -1 \ 0 & -3 & -1 & | & -1 \end{bmatrix} \]
03
Continue Row Reduction
Add the second row to the third row:\[ \begin{bmatrix} 1 & 2 & 2 & | & 1 \ 0 & -3 & -1 & | & -1 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \]This last row indicates that there are infinitely many solutions.
04
Solve for Leading Variables
The second row is rewritten as \(-3x_2 - x_3 = -1\), giving us the equation: \[ x_2 = \frac{1}{3} - \frac{1}{3}x_3 \]First row: \[ x_1 + 2x_2 + 2x_3 = 1 \]Solve for \(x_1\):\[ x_1 = 1 - 2x_2 - 2x_3 \]Substituting for \(x_2\):\[ x_1 = 1 - 2(\frac{1}{3} - \frac{1}{3}x_3) - 2x_3 \]\[ x_1 = \frac{1}{3} - \frac{4}{3}x_3 \]
05
Parametrize and Find Particular Solutions
Parametrize \(x_3 = t\), which results in:\[ x_1 = \frac{1}{3} - \frac{4}{3}t \]\[ x_2 = \frac{1}{3} - \frac{1}{3}t \]\[ x_3 = t \]For \(t=0\), particular solution: \( x_1 = \frac{1}{3}, x_2 = \frac{1}{3}, x_3 = 0 \).For \(t=1\), particular solution: \( x_1 = -1, x_2 = 0, x_3 = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gaussian elimination
Gaussian elimination is a method used to solve systems of linear equations. It involves transforming a matrix into a simpler form, called row-echelon or reduced row-echelon form, which makes it easier to find the solutions. The key goal is to use row operations to create zeros below the pivot positions (usually leading 1s) in a sequence of steps. This reduces complex systems of equations to simpler ones.
- Start by writing the linear system as an augmented matrix.
- Use row operations to simplify. These include swapping rows, multiplying a row by a non-zero number, and adding or subtracting rows.
- Aim to create a leading 1 in each row, if possible.
- Work from the top row to the bottom row, eliminating variables step by step.
row reduction
Row reduction is a process of transforming a matrix into its row-echelon form or reduced row-echelon form through a series of elementary row operations. This process simplifies solving systems of equations and is central to Gaussian elimination.
- Elementary row operations include row swapping, row multiplication, and row addition or subtraction.
- The goal is to manipulate the matrix to achieve as many zeros as possible below the main diagonal, facilitating easy back-substitution.
- During the reduction process, maintain consistency between the system of equations and the matrix.
matrix equation
A matrix equation represents a system of linear equations in a compact form, using matrices and vectors. It fundamentally consists of a coefficient matrix, a variable matrix, and a resultant vector, expressed generally as \(A \vec{x} = \vec{b}\).
- \(A\) is the coefficient matrix, holding the coefficients of the variables in the linear system.
- \(\vec{x}\) is the vector of variables, which you need to solve for.
- \(\vec{b}\) is the resultant vector, representing the constants on the right-hand side of the equations.
infinitely many solutions
A system of linear equations is said to have infinitely many solutions when there are free variables that do not lead to a unique solvable structure. This usually happens when one or more of the equations can be expressed as a combination of others, meaning there is dependence among them.In the transformation to row-echelon form using Gaussian elimination or row reduction, a row of zeros typically indicates infinitely many solutions:
- Free variables arise due to lack of pivot positions that define independent equations.
- The solution can be expressed using parameters that represent the degrees of freedom in the system.
- Specific values for the parameters yield different particular solutions.