Solve The System Using A Matrix:$\[ \begin{cases} 6x + 2y - 2z = 10 \\ -x - 3y + 7z = -27 \\ 3x + 5y - 6z = 18 \\ \end{cases} \\]What Is The Value Of The Unknown Component Represented By [?] In The Solution (1, -3, [?])?

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Introduction

In this article, we will explore the method of solving a system of linear equations using a matrix. This method is a powerful tool for solving systems of linear equations and is widely used in various fields such as physics, engineering, and computer science. We will use the given system of linear equations to demonstrate the method and find the value of the unknown component represented by "[?]" in the solution (1, -3, [?]).

The System of Linear Equations

The given system of linear equations is:

{ \begin{cases} 6x + 2y - 2z = 10 \\ -x - 3y + 7z = -27 \\ 3x + 5y - 6z = 18 \\ \end{cases} \}

Representing the System as a Matrix Equation

We can represent the system of linear equations as a matrix equation in the form:

{ \begin{bmatrix} 6 & 2 & -2 \\ -1 & -3 & 7 \\ 3 & 5 & -6 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 10 \\ -27 \\ 18 \\ \end{bmatrix} \}

Finding the Inverse of the Coefficient Matrix

To solve the system of linear equations, we need to find the inverse of the coefficient matrix. The coefficient matrix is:

{ \begin{bmatrix} 6 & 2 & -2 \\ -1 & -3 & 7 \\ 3 & 5 & -6 \\ \end{bmatrix} \}

We can find the inverse of the coefficient matrix using various methods such as Gaussian elimination or LU decomposition. However, for this example, we will use the method of cofactor expansion.

Calculating the Determinant of the Coefficient Matrix

To find the inverse of the coefficient matrix, we need to calculate the determinant of the coefficient matrix. The determinant of a 3x3 matrix can be calculated using the formula:

{ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \}

where a, b, c, d, e, f, g, h, and i are the elements of the matrix.

Applying the formula to the coefficient matrix, we get:

{ \det(A) = 6((-3)(-6) - (7)(5)) - 2((-1)(-6) - (7)(3)) + (-2)((-1)(5) - (-3)(3)) \}

Simplifying the expression, we get:

{ \det(A) = 6(18 - 35) - 2(6 - 21) + (-2)(-5 - 9) \}

{ \det(A) = 6(-17) - 2(-15) + (-2)(-14) \}

{ \det(A) = -102 + 30 + 28 \}

{ \det(A) = -44 \}

Finding the Adjugate Matrix

The adjugate matrix is the transpose of the cofactor matrix. The cofactor matrix is:

{ \begin{bmatrix} (-3)(-6) - (7)(5) & (-1)(-6) - (7)(3) & (-1)(5) - (-3)(3) \\ (7)(2) - (-2)(5) & (-2)(-2) - (-2)(3) & (2)(5) - (6)(2) \\ (-2)(-3) - (2)(7) & (-2)(-1) - (6)(7) & (2)(-1) - (6)(-3) \\ \end{bmatrix} \}

Simplifying the expression, we get:

{ \begin{bmatrix} 18 - 35 & 6 - 21 & -5 + 9 \\ 14 + 10 & 4 + 6 & 10 - 12 \\ 6 - 14 & -2 - 42 & -2 + 18 \\ \end{bmatrix} \}

{ \begin{bmatrix} -17 & -15 & 4 \\ 24 & 10 & -2 \\ -8 & -44 & 16 \\ \end{bmatrix} \}

The adjugate matrix is the transpose of the cofactor matrix:

{ \begin{bmatrix} -17 & 24 & -8 \\ -15 & 10 & -44 \\ 4 & -2 & 16 \\ \end{bmatrix} \}

Finding the Inverse of the Coefficient Matrix

The inverse of the coefficient matrix is the adjugate matrix divided by the determinant of the coefficient matrix:

{ A^{-1} = \frac{1}{\det(A)} \times \text{adj}(A) \}

Substituting the values, we get:

{ A^{-1} = \frac{1}{-44} \times \begin{bmatrix} -17 & 24 & -8 \\ -15 & 10 & -44 \\ 4 & -2 & 16 \\ \end{bmatrix} \}

Simplifying the expression, we get:

{ A^{-1} = \begin{bmatrix} \frac{17}{44} & -\frac{24}{44} & \frac{8}{44} \\ \frac{15}{44} & -\frac{10}{44} & \frac{44}{44} \\ -\frac{4}{44} & \frac{2}{44} & -\frac{16}{44} \\ \end{bmatrix} \}

{ A^{-1} = \begin{bmatrix} \frac{17}{44} & -\frac{6}{11} & \frac{2}{11} \\ \frac{15}{44} & -\frac{5}{11} & 1 \\ -\frac{1}{11} & \frac{1}{22} & -\frac{4}{11} \\ \end{bmatrix} \}

Solving the System of Linear Equations

Now that we have the inverse of the coefficient matrix, we can solve the system of linear equations by multiplying the inverse of the coefficient matrix with the constant matrix:

{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{17}{44} & -\frac{6}{11} & \frac{2}{11} \\ \frac{15}{44} & -\frac{5}{11} & 1 \\ -\frac{1}{11} & \frac{1}{22} & -\frac{4}{11} \\ \end{bmatrix} \begin{bmatrix} 10 \\ -27 \\ 18 \\ \end{bmatrix} \}

Multiplying the matrices, we get:

{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{17}{44} \times 10 + \frac{6}{11} \times 27 + \frac{2}{11} \times 18 \\ \frac{15}{44} \times 10 + \frac{5}{11} \times 27 + 1 \times 18 \\ -\frac{1}{11} \times 10 + \frac{1}{22} \times 27 + \frac{4}{11} \times 18 \\ \end{bmatrix} \}

Simplifying the expression, we get:

{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{170}{44} + \frac{162}{11} + \frac{36}{11} \\ \frac{150}{44} + \frac{135}{11} + 18 \\ -\frac{10}{11} + \frac{27}{22} + \frac{72}{11} \\ \end{bmatrix} \}

{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{170 + 1458 + 324}{44} \\ \frac{150 + 135 + 792}{44} \\ -\frac{10 + 27 + 72}{22} \\ \end{bmatrix} \}

{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{1952}{44} \\ \frac{1077}{44} \\ -\frac{109}{22} \\ \end{bmatrix} \}

{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 44.5455 \\ 24.4318 \\ -4.9545 \\ \end{bmatrix} \}

Finding the Value of the Unknown Component

The solution to the system of linear equations is (1, -3, [?]). We can find the value of the unknown component represented by "[?]" by comparing the solution with the values we obtained:

{ \begin{bmatrix} x \\ y \\ z \\ <br/> # **Solving a System of Linear Equations using a Matrix: Q&A** =========================================================== ## **Introduction** --------------- In our previous article, we explored the method of solving a system of linear equations using a matrix. We used the given system of linear equations to demonstrate the method and find the value of the unknown component represented by "[?]" in the solution (1, -3, [?]). In this article, we will answer some frequently asked questions related to solving a system of linear equations using a matrix. ## **Q: What is the purpose of using a matrix to solve a system of linear equations?** --------------------------------------------------------- A: The purpose of using a matrix to solve a system of linear equations is to provide a systematic and efficient method for solving systems of linear equations. This method allows us to represent the system of linear equations as a matrix equation and use various techniques such as Gaussian elimination or LU decomposition to solve the system. ## **Q: What is the coefficient matrix in a system of linear equations?** --------------------------------------------------------- A: The coefficient matrix in a system of linear equations is the matrix that contains the coefficients of the variables in the system of linear equations. For example, in the system of linear equations: $\[ \begin{cases} 6x + 2y - 2z = 10 \\ -x - 3y + 7z = -27 \\ 3x + 5y - 6z = 18 \\ \end{cases} \}

The coefficient matrix is:

{ \begin{bmatrix} 6 & 2 & -2 \\ -1 & -3 & 7 \\ 3 & 5 & -6 \\ \end{bmatrix} \}

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

A: To find the inverse of the coefficient matrix, you can use various methods such as Gaussian elimination or LU decomposition. However, for this example, we used the method of cofactor expansion. The inverse of the coefficient matrix is the adjugate matrix divided by the determinant of the coefficient matrix.

Q: What is the adjugate matrix?

A: The adjugate matrix is the transpose of the cofactor matrix. The cofactor matrix is a matrix that contains the cofactors of the elements of the original matrix. The cofactors are obtained by replacing each element of the original matrix with its minor and then multiplying the minor by either 1 or -1 depending on the position of the element.

Q: How do I find the determinant of the coefficient matrix?

A: To find the determinant of the coefficient matrix, you can use various methods such as the formula for the determinant of a 3x3 matrix or the method of cofactor expansion. The determinant of a 3x3 matrix can be calculated using the formula:

{ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \}

where a, b, c, d, e, f, g, h, and i are the elements of the matrix.

Q: What is the solution to the system of linear equations?

A: The solution to the system of linear equations is the values of the variables that satisfy the system of linear equations. In this example, we found the solution to be (1, -3, [?]).

Q: How do I find the value of the unknown component represented by "[?]" in the solution?

A: To find the value of the unknown component represented by "[?]" in the solution, you can compare the solution with the values we obtained:

{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 44.5455 \\ 24.4318 \\ -4.9545 \\ \end{bmatrix} \}

The value of the unknown component represented by "[?]" is -4.9545.

Conclusion

In this article, we answered some frequently asked questions related to solving a system of linear equations using a matrix. We also provided a step-by-step guide on how to solve a system of linear equations using a matrix and find the value of the unknown component represented by "[?]" in the solution. We hope that this article has been helpful in understanding the method of solving a system of linear equations using a matrix.

References

  • [1] "Linear Algebra and Its Applications" by Gilbert Strang
  • [2] "Introduction to Linear Algebra" by Gilbert Strang
  • [3] "Linear Algebra: A Modern Introduction" by David Poole

Further Reading

If you want to learn more about solving systems of linear equations using a matrix, we recommend the following resources:

  • [1] "Linear Algebra and Its Applications" by Gilbert Strang
  • [2] "Introduction to Linear Algebra" by Gilbert Strang
  • [3] "Linear Algebra: A Modern Introduction" by David Poole

We hope that this article has been helpful in understanding the method of solving a system of linear equations using a matrix. If you have any further questions or need help with anything else, please don't hesitate to ask.