Solve The Linear System:$\[ \begin{array}{c} x + 3y - 5z = 7 \\ 4x - 2z = 1 \\ -11x + 3y + Z = -4 \end{array} \\]

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Introduction

A linear system of three equations with three variables is a set of three linear equations in three variables. Solving such a system involves finding the values of the variables that satisfy all three equations simultaneously. In this article, we will discuss the methods for solving a linear system of three equations with three variables.

Methods for Solving a Linear System

There are several methods for solving a linear system of three equations with three variables. Some of the most common methods include:

  • Gaussian Elimination: This method involves transforming the system of equations into upper triangular form using elementary row operations. The system can then be solved by back substitution.
  • Gaussian-Jordan Elimination: This method is similar to Gaussian elimination, but it involves transforming the system into reduced row echelon form.
  • Cramer's Rule: This method involves using determinants to find the values of the variables.
  • Matrix Inversion: This method involves finding the inverse of the coefficient matrix and multiplying it by the constant matrix.

Gaussian Elimination

Gaussian elimination is a method for solving a linear system of three equations with three variables. The method involves transforming the system into upper triangular form using elementary row operations.

Step 1: Write the System of Equations

The system of equations is given by:

{ \begin{array}{c} x + 3y - 5z = 7 \\ 4x - 2z = 1 \\ -11x + 3y + z = -4 \end{array} \}

Step 2: Transform the System into Upper Triangular Form

To transform the system into upper triangular form, we need to perform elementary row operations. The first step is to multiply the first equation by 4 and subtract it from the second equation.

{ \begin{array}{c} x + 3y - 5z = 7 \\ 0x + 12y + 10z = 25 \\ -11x + 3y + z = -4 \end{array} \}

Next, we need to multiply the first equation by 11 and subtract it from the third equation.

{ \begin{array}{c} x + 3y - 5z = 7 \\ 0x + 12y + 10z = 25 \\ 0x + 38y - 54z = 51 \end{array} \}

Step 3: Solve the System by Back Substitution

Now that the system is in upper triangular form, we can solve it by back substitution. We start by solving the third equation for y.

{ y = \frac{51 + 54z}{38} \}

Next, we substitute this expression for y into the second equation and solve for z.

{ 12\left(\frac{51 + 54z}{38}\right) + 10z = 25 \}

Simplifying and solving for z, we get:

{ z = \frac{25 - 612}{190} = -3 \}

Finally, we substitute this value of z into the first equation and solve for x.

{ x + 3\left(\frac{51 + 54(-3)}{38}\right) - 5(-3) = 7 \}

Simplifying and solving for x, we get:

{ x = 2 \}

Therefore, the solution to the system is x = 2, y = 1, and z = -3.

Gaussian-Jordan Elimination

Gaussian-Jordan elimination is a method for solving a linear system of three equations with three variables. The method involves transforming the system into reduced row echelon form.

Step 1: Write the System of Equations

The system of equations is given by:

{ \begin{array}{c} x + 3y - 5z = 7 \\ 4x - 2z = 1 \\ -11x + 3y + z = -4 \end{array} \}

Step 2: Transform the System into Reduced Row Echelon Form

To transform the system into reduced row echelon form, we need to perform elementary row operations. The first step is to multiply the first equation by 4 and subtract it from the second equation.

{ \begin{array}{c} x + 3y - 5z = 7 \\ 0x + 12y + 10z = 25 \\ -11x + 3y + z = -4 \end{array} \}

Next, we need to multiply the first equation by 11 and subtract it from the third equation.

{ \begin{array}{c} x + 3y - 5z = 7 \\ 0x + 12y + 10z = 25 \\ 0x + 38y - 54z = 51 \end{array} \}

Step 3: Solve the System by Back Substitution

Now that the system is in reduced row echelon form, we can solve it by back substitution. We start by solving the third equation for y.

{ y = \frac{51 + 54z}{38} \}

Next, we substitute this expression for y into the second equation and solve for z.

{ 12\left(\frac{51 + 54z}{38}\right) + 10z = 25 \}

Simplifying and solving for z, we get:

{ z = \frac{25 - 612}{190} = -3 \}

Finally, we substitute this value of z into the first equation and solve for x.

{ x + 3\left(\frac{51 + 54(-3)}{38}\right) - 5(-3) = 7 \}

Simplifying and solving for x, we get:

{ x = 2 \}

Therefore, the solution to the system is x = 2, y = 1, and z = -3.

Cramer's Rule

Cramer's rule is a method for solving a linear system of three equations with three variables. The method involves using determinants to find the values of the variables.

Step 1: Find the Determinant of the Coefficient Matrix

The coefficient matrix is given by:

{ \begin{array}{c} 1 & 3 & -5 \\ 4 & 0 & -2 \\ -11 & 3 & 1 \end{array} \}

The determinant of the coefficient matrix is:

{ \begin{vmatrix} 1 & 3 & -5 \\ 4 & 0 & -2 \\ -11 & 3 & 1 \end{vmatrix} \}

Expanding the determinant along the first row, we get:

{ 1\begin{vmatrix} 0 & -2 \\ 3 & 1 \end{vmatrix} - 3\begin{vmatrix} 4 & -2 \\ -11 & 1 \end{vmatrix} + (-5)\begin{vmatrix} 4 & 0 \\ -11 & 3 \end{vmatrix} \}

Simplifying and evaluating the determinants, we get:

{ 1(0 + 6) - 3(4 + 22) - 5(12 + 0) \}

{ 6 - 78 - 60 \}

{ -132 \}

Step 2: Find the Determinant of the Coefficient Matrix with x Replaced by 1

The coefficient matrix with x replaced by 1 is given by:

{ \begin{array}{c} 1 & 3 & -5 \\ 4 & 0 & -2 \\ -11 & 3 & 1 \end{array} \}

The determinant of the coefficient matrix with x replaced by 1 is:

{ \begin{vmatrix} 1 & 3 & -5 \\ 4 & 0 & -2 \\ -11 & 3 & 1 \end{vmatrix} \}

Expanding the determinant along the first row, we get:

{ 1\begin{vmatrix} 0 & -2 \\ 3 & 1 \end{vmatrix} - 3\begin{vmatrix} 4 & -2 \\ -11 & 1 \end{vmatrix} + (-5)\begin{vmatrix} 4 & 0 \\ -11 & 3 \end{vmatrix} \}

Simplifying and evaluating the determinants, we get:

{ 1(0 + 6) - 3(4 + 22) - 5(12 + 0) \}

{ 6 - 78 - 60 \}

{ -132 \}

Step 3: Find the Determinant of the Coefficient Matrix with y Replaced by 1

The coefficient matrix with y replaced by 1 is given by:

{ \begin{array}{c} 1 & 3 & -5 \\ 4 & 0 & -2 \\ -11 & 3 & 1 \end{array} \}

The determinant of the coefficient matrix with y replaced by 1 is:

{ \begin{vmatrix} 1 & 3 & -5 \\ 4 & 0 & -2 \\ -11 & 3 & 1 \end{vmatrix} \}

Q: What is a linear system of three equations with three variables?

A: A linear system of three equations with three variables is a set of three linear equations in three variables. Solving such a system involves finding the values of the variables that satisfy all three equations simultaneously.

Q: What are some common methods for solving a linear system of three equations with three variables?

A: Some common methods for solving a linear system of three equations with three variables include:

  • Gaussian Elimination: This method involves transforming the system of equations into upper triangular form using elementary row operations. The system can then be solved by back substitution.
  • Gaussian-Jordan Elimination: This method is similar to Gaussian elimination, but it involves transforming the system into reduced row echelon form.
  • Cramer's Rule: This method involves using determinants to find the values of the variables.
  • Matrix Inversion: This method involves finding the inverse of the coefficient matrix and multiplying it by the constant matrix.

Q: How do I use Gaussian Elimination to solve a linear system of three equations with three variables?

A: To use Gaussian elimination to solve a linear system of three equations with three variables, follow these steps:

  1. Write the system of equations in the form Ax = b, where A is the coefficient matrix, x is the variable matrix, and b is the constant matrix.
  2. Perform elementary row operations to transform the system into upper triangular form.
  3. Solve the system by back substitution.

Q: How do I use Gaussian-Jordan Elimination to solve a linear system of three equations with three variables?

A: To use Gaussian-Jordan elimination to solve a linear system of three equations with three variables, follow these steps:

  1. Write the system of equations in the form Ax = b, where A is the coefficient matrix, x is the variable matrix, and b is the constant matrix.
  2. Perform elementary row operations to transform the system into reduced row echelon form.
  3. Solve the system by back substitution.

Q: How do I use Cramer's Rule to solve a linear system of three equations with three variables?

A: To use Cramer's rule to solve a linear system of three equations with three variables, follow these steps:

  1. Find the determinant of the coefficient matrix.
  2. Find the determinant of the coefficient matrix with x replaced by 1.
  3. Find the determinant of the coefficient matrix with y replaced by 1.
  4. Find the determinant of the coefficient matrix with z replaced by 1.
  5. Use the determinants to find the values of x, y, and z.

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

A: Some common mistakes to avoid when solving a linear system of three equations with three variables include:

  • Not following the correct order of operations: Make sure to follow the correct order of operations when performing elementary row operations.
  • Not checking for consistency: Make sure to check for consistency between the equations and the variables.
  • Not using the correct method: Make sure to use the correct method for solving the system, such as Gaussian elimination or Cramer's rule.

Q: How do I check if a linear system of three equations with three variables has a unique solution?

A: To check if a linear system of three equations with three variables has a unique solution, follow these steps:

  1. Find the determinant of the coefficient matrix.
  2. If the determinant is non-zero, the system has a unique solution.
  3. If the determinant is zero, the system may have no solution or infinitely many solutions.

Q: How do I check if a linear system of three equations with three variables has no solution?

A: To check if a linear system of three equations with three variables has no solution, follow these steps:

  1. Find the determinant of the coefficient matrix.
  2. If the determinant is zero, the system may have no solution or infinitely many solutions.
  3. Check if the system is inconsistent by checking if the equations are contradictory.

Q: How do I check if a linear system of three equations with three variables has infinitely many solutions?

A: To check if a linear system of three equations with three variables has infinitely many solutions, follow these steps:

  1. Find the determinant of the coefficient matrix.
  2. If the determinant is zero, the system may have no solution or infinitely many solutions.
  3. Check if the system is consistent by checking if the equations are consistent.

Conclusion

Solving a linear system of three equations with three variables involves finding the values of the variables that satisfy all three equations simultaneously. There are several methods for solving such a system, including Gaussian elimination, Gaussian-Jordan elimination, Cramer's rule, and matrix inversion. By following the correct steps and avoiding common mistakes, you can solve a linear system of three equations with three variables and find the values of the variables.