Solve The System. If There Is No Solution Or If There Are Infinitely Many Solutions And The System's Equations Are Dependent, State So.${ \begin{array}{rr} 6x - Y + 2z = & -23 \ x + 2y - Z = & 4 \ 2x + 2y - 3z = & 3 \end{array} }$

by ADMIN 232 views

=====================================================

Introduction

Solving a system of linear equations involves finding the values of variables that satisfy all the equations in the system. In this article, we will explore how to solve a system of linear equations using the method of substitution and elimination. We will also discuss how to determine if a system has no solution or infinitely many solutions.

The System of Linear Equations

The system of linear equations we will be solving is:

{ \begin{array}{rr} 6x - y + 2z = & -23 \\ x + 2y - z = & 4 \\ 2x + 2y - 3z = & 3 \end{array} \}

Step 1: Write Down the Augmented Matrix

To solve the system of linear equations, we will first write down the augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constants on the right-hand side of the equations.

{ \left[ \begin{array}{rrr|r} 6 & -1 & 2 & -23 \\ 1 & 2 & -1 & 4 \\ 2 & 2 & -3 & 3 \end{array} \right] \}

Step 2: Perform Row Operations

To solve the system of linear equations, we will perform row operations on the augmented matrix. The goal is to transform the matrix into row-echelon form, where all the entries below the leading entry in each row are zeros.

Step 2.1: Multiply Row 1 by 1/6

We will multiply row 1 by 1/6 to get a leading entry of 1.

{ \left[ \begin{array}{rrr|r} 1 & -1/6 & 1/3 & -23/6 \\ 1 & 2 & -1 & 4 \\ 2 & 2 & -3 & 3 \end{array} \right] \}

Step 2.2: Subtract Row 1 from Row 2

We will subtract row 1 from row 2 to get a zero in the first column.

{ \left[ \begin{array}{rrr|r} 1 & -1/6 & 1/3 & -23/6 \\ 0 & 13/6 & -4/3 & 41/6 \\ 2 & 2 & -3 & 3 \end{array} \right] \}

Step 2.3: Subtract 2 Times Row 1 from Row 3

We will subtract 2 times row 1 from row 3 to get a zero in the first column.

{ \left[ \begin{array}{rrr|r} 1 & -1/6 & 1/3 & -23/6 \\ 0 & 13/6 & -4/3 & 41/6 \\ 0 & 7/3 & -5/3 & 41/3 \end{array} \right] \}

Step 2.4: Multiply Row 2 by 6/13

We will multiply row 2 by 6/13 to get a leading entry of 1.

{ \left[ \begin{array}{rrr|r} 1 & -1/6 & 1/3 & -23/6 \\ 0 & 1 & -4/13 & 41/13 \\ 0 & 7/3 & -5/3 & 41/3 \end{array} \right] \}

Step 2.5: Subtract 7/3 Times Row 2 from Row 3

We will subtract 7/3 times row 2 from row 3 to get a zero in the second column.

{ \left[ \begin{array}{rrr|r} 1 & -1/6 & 1/3 & -23/6 \\ 0 & 1 & -4/13 & 41/13 \\ 0 & 0 & -1/39 & -6/13 \end{array} \right] \}

Step 2.6: Multiply Row 3 by -39

We will multiply row 3 by -39 to get a leading entry of 1.

{ \left[ \begin{array}{rrr|r} 1 & -1/6 & 1/3 & -23/6 \\ 0 & 1 & -4/13 & 41/13 \\ 0 & 0 & 1 & 6 \end{array} \right] \}

Step 3: Back-Substitution

Now that we have the matrix in row-echelon form, we can perform back-substitution to find the values of the variables.

Step 3.1: Solve for z

We will solve for z by using the last row of the matrix.

z=6z = 6

Step 3.2: Solve for y

We will solve for y by using the second row of the matrix.

y−413z=4113y - \frac{4}{13}z = \frac{41}{13}

y=4113+413⋅6y = \frac{41}{13} + \frac{4}{13} \cdot 6

y=4113+2413y = \frac{41}{13} + \frac{24}{13}

y=6513y = \frac{65}{13}

Step 3.3: Solve for x

We will solve for x by using the first row of the matrix.

x−16y+13z=−236x - \frac{1}{6}y + \frac{1}{3}z = -\frac{23}{6}

x=−236+16y−13zx = -\frac{23}{6} + \frac{1}{6}y - \frac{1}{3}z

x=−236+16⋅6513−13⋅6x = -\frac{23}{6} + \frac{1}{6} \cdot \frac{65}{13} - \frac{1}{3} \cdot 6

x=−236+6578−2x = -\frac{23}{6} + \frac{65}{78} - 2

x=−236+6578−15678x = -\frac{23}{6} + \frac{65}{78} - \frac{156}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178⋅66x = -\frac{23}{6} - \frac{91}{78} \cdot \frac{6}{6}

x=−236−9178⋅66x = -\frac{23}{6} - \frac{91}{78} \cdot \frac{6}{6}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

x=−236−9178x = -\frac{23}{6} - \frac{91}{78}

$x = -\frac{23}{6

=====================================================

Introduction

Solving a system of linear equations can be a challenging task, but with the right approach, it can be made easier. In this article, we will provide a Q&A section to help you understand the process of solving a system of linear equations.

Q: What is a system of linear equations?

A system of linear equations is a set of two or more linear equations that are related to each other through a common variable or variables.

A: How do I know if a system of linear equations has a solution?

A system of linear equations has a solution if the number of equations is equal to the number of variables. If the number of equations is greater than the number of variables, the system may have no solution or infinitely many solutions.

Q: What is the difference between a solution and an infinitely many solutions?

A solution is a specific set of values that satisfy all the equations in the system. Infinitely many solutions, on the other hand, means that there are an infinite number of possible solutions that satisfy all the equations in the system.

A: How do I determine if a system of linear equations has infinitely many solutions?

A system of linear equations has infinitely many solutions if the number of equations is greater than the number of variables, and the equations are linearly dependent. Linearly dependent equations are equations that can be expressed as a linear combination of other equations.

Q: What is the method of substitution?

The method of substitution is a technique used to solve a system of linear equations by substituting the value of one variable into another equation.

A: What is the method of elimination?

The method of elimination is a technique used to solve a system of linear equations by adding or subtracting equations to eliminate one or more variables.

Q: What is the augmented matrix?

The augmented matrix is a matrix that includes the coefficients of the variables and the constants on the right-hand side of the equations.

A: How do I perform row operations on the augmented matrix?

To perform row operations on the augmented matrix, you need to follow these steps:

  1. Multiply a row by a non-zero constant.
  2. Add a multiple of one row to another row.
  3. Subtract a multiple of one row from another row.

Q: What is row-echelon form?

Row-echelon form is a matrix that has all the entries below the leading entry in each row equal to zero.

A: How do I perform back-substitution?

To perform back-substitution, you need to follow these steps:

  1. Solve for the last variable using the last row of the matrix.
  2. Substitute the value of the last variable into the second-to-last row of the matrix.
  3. Solve for the second-to-last variable using the second-to-last row of the matrix.
  4. Repeat the process until all variables have been solved.

Q: What are some common mistakes to avoid when solving a system of linear equations?

Some common mistakes to avoid when solving a system of linear equations include:

  • Not following the order of operations.
  • Not checking for linear dependence between equations.
  • Not performing row operations correctly.
  • Not performing back-substitution correctly.

A: How do I check if a system of linear equations has no solution?

To check if a system of linear equations has no solution, you need to follow these steps:

  1. Check if the number of equations is greater than the number of variables.
  2. Check if the equations are linearly dependent.
  3. If the equations are linearly dependent, check if the system has infinitely many solutions.

Q: What are some real-world applications of solving a system of linear equations?

Some real-world applications of solving a system of linear equations include:

  • Solving a system of linear equations can be used to find the optimal solution to a problem.
  • Solving a system of linear equations can be used to find the minimum or maximum value of a function.
  • Solving a system of linear equations can be used to find the solution to a system of equations that represents a real-world problem.

A: How do I use technology to solve a system of linear equations?

You can use technology such as calculators, computers, or software to solve a system of linear equations. Some popular software for solving systems of linear equations include:

  • MATLAB
  • Mathematica
  • Maple
  • Python

Q: What are some tips for solving a system of linear equations?

Some tips for solving a system of linear equations include:

  • Read the problem carefully and understand what is being asked.
  • Use the method of substitution or elimination to solve the system.
  • Check for linear dependence between equations.
  • Perform row operations correctly.
  • Perform back-substitution correctly.
  • Use technology to check your solution.

A: How do I know if I have solved the system correctly?

To know if you have solved the system correctly, you need to follow these steps:

  1. Check if the solution satisfies all the equations in the system.
  2. Check if the solution is consistent with the problem.
  3. Check if the solution is unique or if there are infinitely many solutions.

By following these tips and using the correct techniques, you can solve a system of linear equations and find the solution to a real-world problem.