What Is The Solution To The System Of Equations?$\[ \begin{aligned} x - Y - 2z &= -2 \\ -x - Y + Z &= 5 \\ 3x + 2y - Z &= -1 \end{aligned} \\]A. \[$(4, -4, 5)\$\]B. \[$(4, -4, -5)\$\]C. \[$(-4, -4, 5)\$\]D. \[$(4,

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Introduction

Solving a system of linear equations is a fundamental concept in mathematics, particularly in algebra and linear algebra. It involves finding the values of variables that satisfy all the equations in the system. In this article, we will explore the solution to a system of three linear equations with three variables. We will use the method of substitution and elimination to find the solution.

The System of Equations

The given system of equations is:

xāˆ’yāˆ’2z=āˆ’2āˆ’xāˆ’y+z=53x+2yāˆ’z=āˆ’1\begin{aligned} x - y - 2z &= -2 \\ -x - y + z &= 5 \\ 3x + 2y - z &= -1 \end{aligned}

Step 1: Write the System of Equations in Matrix Form

To solve the system of equations, we can write it in matrix form as follows:

[1āˆ’1āˆ’2āˆ’1āˆ’1132āˆ’1][xyz]=[āˆ’25āˆ’1]\begin{bmatrix} 1 & -1 & -2 \\ -1 & -1 & 1 \\ 3 & 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 \\ 5 \\ -1 \end{bmatrix}

Step 2: Find the Augmented Matrix

The augmented matrix is obtained by appending the constant matrix to the coefficient matrix:

[1āˆ’1āˆ’2āˆ’2āˆ’1āˆ’11532āˆ’1āˆ’1]\begin{bmatrix} 1 & -1 & -2 & -2 \\ -1 & -1 & 1 & 5 \\ 3 & 2 & -1 & -1 \end{bmatrix}

Step 3: Perform Row Operations to Get the Reduced Row Echelon Form (RREF)

To solve the system of equations, we need to perform row operations to get the reduced row echelon form (RREF) of the augmented matrix.

Step 3.1: Multiply Row 1 by 1 and Add to Row 2

[1āˆ’1āˆ’2āˆ’200āˆ’3732āˆ’1āˆ’1]\begin{bmatrix} 1 & -1 & -2 & -2 \\ 0 & 0 & -3 & 7 \\ 3 & 2 & -1 & -1 \end{bmatrix}

Step 3.2: Multiply Row 1 by -3 and Add to Row 3

[1āˆ’1āˆ’2āˆ’200āˆ’370557]\begin{bmatrix} 1 & -1 & -2 & -2 \\ 0 & 0 & -3 & 7 \\ 0 & 5 & 5 & 7 \end{bmatrix}

Step 3.3: Multiply Row 2 by -1/3

[1āˆ’1āˆ’2āˆ’2001āˆ’7/30557]\begin{bmatrix} 1 & -1 & -2 & -2 \\ 0 & 0 & 1 & -7/3 \\ 0 & 5 & 5 & 7 \end{bmatrix}

Step 3.4: Multiply Row 2 by -5 and Add to Row 3

[1āˆ’1āˆ’2āˆ’2001āˆ’7/300āˆ’200]\begin{bmatrix} 1 & -1 & -2 & -2 \\ 0 & 0 & 1 & -7/3 \\ 0 & 0 & -20 & 0 \end{bmatrix}

Step 3.5: Multiply Row 2 by 20 and Add to Row 3

[1āˆ’1āˆ’2āˆ’2001āˆ’7/30000]\begin{bmatrix} 1 & -1 & -2 & -2 \\ 0 & 0 & 1 & -7/3 \\ 0 & 0 & 0 & 0 \end{bmatrix}

Step 4: Interpret the RREF

The RREF of the augmented matrix is:

[1āˆ’1āˆ’2āˆ’2001āˆ’7/30000]\begin{bmatrix} 1 & -1 & -2 & -2 \\ 0 & 0 & 1 & -7/3 \\ 0 & 0 & 0 & 0 \end{bmatrix}

This means that the system of equations has infinitely many solutions. We can express the solution in terms of a parameter, say tt.

Step 5: Express the Solution in Terms of a Parameter

Let z=tz = t. Then, we have:

y=āˆ’7/3+3ty = -7/3 + 3t

x=āˆ’2+y+2z=āˆ’2+(āˆ’7/3+3t)+2t=āˆ’2āˆ’7/3+5tx = -2 + y + 2z = -2 + (-7/3 + 3t) + 2t = -2 - 7/3 + 5t

Step 6: Simplify the Solution

Simplifying the solution, we get:

x=āˆ’11/3+5tx = -11/3 + 5t

y=āˆ’7/3+3ty = -7/3 + 3t

z=tz = t

Step 7: Find the Particular Solution

To find the particular solution, we need to find the value of tt that satisfies the system of equations.

Substituting x=4x = 4, y=āˆ’4y = -4, and z=5z = 5 into the system of equations, we get:

4āˆ’(āˆ’4)āˆ’2(5)=āˆ’24 - (-4) - 2(5) = -2

āˆ’4āˆ’(āˆ’4)+5=5-4 - (-4) + 5 = 5

3(4)+2(āˆ’4)āˆ’5=āˆ’13(4) + 2(-4) - 5 = -1

All three equations are satisfied, so the particular solution is:

x=4x = 4

y=āˆ’4y = -4

z=5z = 5

Conclusion

In conclusion, the solution to the system of equations is:

(4,āˆ’4,5)\boxed{(4, -4, 5)}

This solution satisfies all three equations in the system.

Introduction

In the previous article, we solved a system of three linear equations with three variables using the method of substitution and elimination. In this article, we will answer some frequently asked questions (FAQs) about the system of equations.

Q: What is a system of equations?

A: A system of equations is a set of two or more equations that involve two or more variables. In this article, we solved a system of three linear equations with three variables.

Q: What is the difference between a system of equations and a single equation?

A: A single equation is a statement that expresses a relationship between two or more variables. A system of equations, on the other hand, is a set of two or more equations that involve two or more variables.

Q: How do I know if a system of equations has a unique solution, infinitely many solutions, or no solution?

A: To determine the number of solutions to a system of equations, you can use the following methods:

  • If the system of equations has a unique solution, the augmented matrix will have a unique solution.
  • If the system of equations has infinitely many solutions, the augmented matrix will have a row of zeros with a non-zero constant term.
  • If the system of equations has no solution, the augmented matrix will have a row of zeros with a zero constant term.

Q: How do I solve a system of equations using the method of substitution?

A: To solve a system of equations using the method of substitution, follow these steps:

  1. Choose one of the equations and solve it for one of the variables.
  2. Substitute the expression for the variable into the other equation.
  3. Solve the resulting equation for the other variable.
  4. Substitute the value of the other variable back into one of the original equations to find the value of the first variable.

Q: How do I solve a system of equations using the method of elimination?

A: To solve a system of equations using the method of elimination, follow these steps:

  1. Choose two of the equations and multiply them by necessary multiples such that the coefficients of one of the variables are the same.
  2. Add the two equations to eliminate the variable.
  3. Solve the resulting equation for the other variable.
  4. Substitute the value of the other variable back into one of the original equations to find the value of the first variable.

Q: What is the reduced row echelon form (RREF) of a matrix?

A: The reduced row echelon form (RREF) of a matrix is a matrix that has the following properties:

  • All the rows that contain only zeros are at the bottom of the matrix.
  • Each row that is not all zeros has a 1 as its first nonzero entry (this entry is called a leading 1).
  • The column in which a leading 1 of a row is found has all zeros elsewhere, so a column containing a leading 1 will have zeros everywhere except for one place.

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

A: To find the RREF of a matrix, follow these steps:

  1. Start with the original matrix.
  2. Perform row operations to get the matrix into row echelon form.
  3. Perform additional row operations to get the matrix into reduced row echelon form.

Q: What is the difference between row echelon form and reduced row echelon form?

A: Row echelon form is a matrix that has the following properties:

  • All the rows that contain only zeros are at the bottom of the matrix.
  • Each row that is not all zeros has a 1 as its first nonzero entry (this entry is called a leading 1).
  • The column in which a leading 1 of a row is found has all zeros elsewhere, so a column containing a leading 1 will have zeros everywhere except for one place.

Reduced row echelon form is a matrix that has the same properties as row echelon form, but with the additional property that each leading 1 is to the right of the leading 1 in the row above it.

Q: How do I use the RREF to solve a system of equations?

A: To use the RREF to solve a system of equations, follow these steps:

  1. Write the system of equations in matrix form.
  2. Find the RREF of the matrix.
  3. Interpret the RREF to find the solution to the system of equations.

Conclusion

In conclusion, we have answered some frequently asked questions (FAQs) about the system of equations. We hope that this article has been helpful in clarifying any confusion you may have had about the system of equations.