Solve The Following System Of Equations Using A Matrix Equation. Check Your Answers.$\[ \begin{aligned} 9y + 2z &= 23 \\ 3x + 2y + Z &= 7 \\ x - Y &= -2 \end{aligned} \\]Select The Correct Choice And, If Necessary, Fill In The Answer Boxes To

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Introduction

In mathematics, a system of equations is a set of equations that involve multiple variables. Solving a system of equations involves finding the values of the variables that satisfy all the equations simultaneously. In this article, we will discuss how to solve a system of equations using matrix equations.

What is a Matrix Equation?

A matrix equation is an equation that involves matrices and vectors. Matrices are rectangular arrays of numbers, and vectors are matrices with a single column or row. Matrix equations are used to solve systems of linear equations, and they have many applications in mathematics, science, and engineering.

Converting a System of Equations to a Matrix Equation

To convert a system of equations to a matrix equation, we need to follow these steps:

  1. Write the system of equations in the form of AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
  2. The coefficient matrix A is obtained by writing the coefficients of the variables in the system of equations as a matrix.
  3. The variable matrix X is obtained by writing the variables in the system of equations as a matrix.
  4. The constant matrix B is obtained by writing the constants in the system of equations as a matrix.

Converting the Given System of Equations to a Matrix Equation

The given system of equations is:

{ \begin{aligned} 9y + 2z &= 23 \\ 3x + 2y + z &= 7 \\ x - y &= -2 \end{aligned} \}

To convert this system of equations to a matrix equation, we need to follow the steps mentioned above.

Step 1: Write the System of Equations in the Form of AX = B

The coefficient matrix A is:

{ \begin{aligned} A = \begin{bmatrix} 0 & 9 & 2 \\ 3 & 2 & 1 \\ 1 & -1 & 0 \end{bmatrix} \end{aligned} \}

The variable matrix X is:

{ \begin{aligned} X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \end{aligned} \}

The constant matrix B is:

{ \begin{aligned} B = \begin{bmatrix} 23 \\ 7 \\ -2 \end{bmatrix} \end{aligned} \}

Step 2: Write the Matrix Equation

The matrix equation is:

{ \begin{aligned} AX = B \end{aligned} \}

{ \begin{aligned} \begin{bmatrix} 0 & 9 & 2 \\ 3 & 2 & 1 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 23 \\ 7 \\ -2 \end{bmatrix} \end{aligned} \}

Solving the Matrix Equation

To solve the matrix equation, we need to find the values of the variables x, y, and z that satisfy the equation.

Method 1: Using the Inverse of the Coefficient Matrix

The inverse of the coefficient matrix A is:

{ \begin{aligned} A^{-1} = \begin{bmatrix} -1/3 & 1/3 & 1/3 \\ -1/3 & 1/3 & -1/3 \\ 1/3 & -1/3 & 1/3 \end{bmatrix} \end{aligned} \}

The solution to the matrix equation is:

{ \begin{aligned} X = A^{-1}B \end{aligned} \}

{ \begin{aligned} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1/3 & 1/3 & 1/3 \\ -1/3 & 1/3 & -1/3 \\ 1/3 & -1/3 & 1/3 \end{bmatrix} \begin{bmatrix} 23 \\ 7 \\ -2 \end{bmatrix} \end{aligned} \}

{ \begin{aligned} X = \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} \end{aligned} \}

Method 2: Using Gaussian Elimination

Gaussian elimination is a method for solving systems of linear equations. It involves transforming the coefficient matrix into upper triangular form using elementary row operations.

Step 1: Transform the Coefficient Matrix into Upper Triangular Form

The coefficient matrix A is:

{ \begin{aligned} A = \begin{bmatrix} 0 & 9 & 2 \\ 3 & 2 & 1 \\ 1 & -1 & 0 \end{bmatrix} \end{aligned} \}

Using elementary row operations, we can transform the coefficient matrix into upper triangular form:

{ \begin{aligned} A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 9 & 2 \\ 3 & 2 & 1 \end{bmatrix} \end{aligned} \}

{ \begin{aligned} A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 9 & 2 \\ 0 & 5 & 1 \end{bmatrix} \end{aligned} \}

{ \begin{aligned} A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 9 & 2 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned} \}

Step 2: Solve the System of Equations

The system of equations is:

{ \begin{aligned} x - y &= -2 \\ 9y + 2z &= 23 \\ x &= 1 \end{aligned} \}

Solving the system of equations, we get:

{ \begin{aligned} x &= 1 \\ y &= 3 \\ z &= -1 \end{aligned} \}

Conclusion

In this article, we discussed how to solve a system of equations using matrix equations. We converted the given system of equations to a matrix equation and solved it using two methods: the inverse of the coefficient matrix and Gaussian elimination. The solution to the system of equations is x = 1, y = 3, and z = -1.

References

  • [1] Strang, G. (1988). Linear Algebra and Its Applications. 3rd ed. Harcourt Brace Jovanovich.
  • [2] Hoffman, K., and Kunze, R. (1971). Linear Algebra. 2nd ed. Prentice Hall.
  • [3] Gantmacher, F. R. (1959). The Theory of Matrices. 2nd ed. Chelsea Publishing Company.
    Frequently Asked Questions (FAQs) about Solving Systems of Equations using Matrix Equations =============================================================================================

Q: What is a matrix equation?

A: A matrix equation is an equation that involves matrices and vectors. Matrices are rectangular arrays of numbers, and vectors are matrices with a single column or row. Matrix equations are used to solve systems of linear equations, and they have many applications in mathematics, science, and engineering.

Q: How do I convert a system of equations to a matrix equation?

A: To convert a system of equations to a matrix equation, you need to follow these steps:

  1. Write the system of equations in the form of AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
  2. The coefficient matrix A is obtained by writing the coefficients of the variables in the system of equations as a matrix.
  3. The variable matrix X is obtained by writing the variables in the system of equations as a matrix.
  4. The constant matrix B is obtained by writing the constants in the system of equations as a matrix.

Q: What is the inverse of a matrix?

A: The inverse of a matrix A is a matrix A^-1 such that AA^-1 = A^-1A = I, where I is the identity matrix. The inverse of a matrix is used to solve systems of linear equations.

Q: How do I find the inverse of a matrix?

A: There are several methods for finding the inverse of a matrix, including:

  1. Gaussian elimination: This method involves transforming the matrix into upper triangular form using elementary row operations.
  2. Determinant method: This method involves finding the determinant of the matrix and then using the formula for the inverse of a matrix.
  3. Adjugate method: This method involves finding the adjugate of the matrix and then dividing it by the determinant of the matrix.

Q: What is Gaussian elimination?

A: Gaussian elimination is a method for solving systems of linear equations. It involves transforming the coefficient matrix into upper triangular form using elementary row operations.

Q: What are elementary row operations?

A: Elementary row operations are operations that can be performed on a matrix to transform it into a different form. These operations include:

  1. Row interchange: This involves swapping two rows of the matrix.
  2. Row multiplication: This involves multiplying a row of the matrix by a scalar.
  3. Row addition: This involves adding a multiple of one row to another row.

Q: How do I use Gaussian elimination to solve a system of equations?

A: To use Gaussian elimination to solve a system of equations, you need to follow these steps:

  1. Write the system of equations in the form of AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
  2. Transform the coefficient matrix into upper triangular form using elementary row operations.
  3. Solve the system of equations by back substitution.

Q: What is back substitution?

A: Back substitution is a method for solving systems of linear equations. It involves solving the system of equations by substituting the values of the variables from the last equation to the first equation.

Q: How do I use back substitution to solve a system of equations?

A: To use back substitution to solve a system of equations, you need to follow these steps:

  1. Write the system of equations in the form of AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
  2. Solve the system of equations by substituting the values of the variables from the last equation to the first equation.

Q: What are the advantages of using matrix equations to solve systems of equations?

A: The advantages of using matrix equations to solve systems of equations include:

  1. Efficiency: Matrix equations can be solved more efficiently than systems of equations.
  2. Accuracy: Matrix equations can be solved with greater accuracy than systems of equations.
  3. Flexibility: Matrix equations can be used to solve a wide range of systems of equations.

Q: What are the disadvantages of using matrix equations to solve systems of equations?

A: The disadvantages of using matrix equations to solve systems of equations include:

  1. Complexity: Matrix equations can be more complex than systems of equations.
  2. Difficulty: Matrix equations can be more difficult to solve than systems of equations.
  3. Computational requirements: Matrix equations require more computational resources than systems of equations.