Math Problem Statement
Solve the following system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. left brace Start 3 By 1 Matrix 1st Row 1st Column 2nd Row 1st Column 3rd Row 1st Column EndMatrix
negative x plus y plus z equals negative 1 negative x plus 4 y minus 14 z equals negative 22 7 x minus 6 y minus 12 z equals 0 Question content area bottom Part 1 Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The solution is left parenthesis nothing comma nothing comma nothing right parenthesis . (Simplify your answers.) B. There are infinitely many solutions. The solution can be written as StartSet left parenthesis x comma y comma z right parenthesis font size decreased by 6 | font size decreased by 6 x equals nothing comma y equals nothing comma z is any real number EndSet . (Simplify your answers. Type expressions using z as the variable.) C. There are infinitely many solutions. The solution can be written as StartSet left parenthesis x comma y comma z right parenthesis font size decreased by 6 | font size decreased by 6 x equals nothing comma y is any real number comma z is any real number EndSet . (Simplify your answer. Type an expression using y and z as the variables.) D. The system is inconsistent.
Solution
To solve this system of equations using matrices and row operations, we represent the system as an augmented matrix, perform row operations to bring it to reduced row-echelon form, and analyze whether the system has a solution.
The system of equations:
Step 1: Write the augmented matrix
-1 & 1 & 1 & -1 \\ -1 & 4 & -14 & -22 \\ 7 & -6 & -12 & 0 \end{bmatrix}$$ We will now perform Gaussian elimination to simplify this. --- ### Step 2: Apply row operations to reduce the matrix #### Step 2.1: Make the pivot at (1,1) a 1 Divide the first row by $$-1$$: $$\begin{bmatrix} 1 & -1 & -1 & 1 \\ -1 & 4 & -14 & -22 \\ 7 & -6 & -12 & 0 \end{bmatrix}$$ #### Step 2.2: Eliminate the first column below the pivot Add $$R_1$$ to $$R_2$$ and subtract $$7R_1$$ from $$R_3$$: $$R_2 \to R_2 + R_1, \quad R_3 \to R_3 - 7R_1$$ $$\begin{bmatrix} 1 & -1 & -1 & 1 \\ 0 & 3 & -15 & -21 \\ 0 & 1 & -5 & -7 \end{bmatrix}$$ #### Step 2.3: Make the pivot at (2,2) a 1 Divide the second row by $$3$$: $$R_2 \to \frac{R_2}{3}$$ $$\begin{bmatrix} 1 & -1 & -1 & 1 \\ 0 & 1 & -5 & -7 \\ 0 & 1 & -5 & -7 \end{bmatrix}$$ #### Step 2.4: Eliminate the second column below and above the pivot Subtract $$R_2$$ from $$R_3$$ and add $$R_2$$ to $$R_1$$: $$R_3 \to R_3 - R_2, \quad R_1 \to R_1 + R_2$$ $$\begin{bmatrix} 1 & 0 & -6 & -6 \\ 0 & 1 & -5 & -7 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ --- ### Step 3: Analyze the reduced matrix The third row is $$0 = 0$$, so it is consistent. However, the system has infinitely many solutions because the third variable ($$z$$) is free. ### Step 4: Solve for the variables From the reduced matrix: 1. $$x - 6z = -6 \implies x = -6 + 6z$$ 2. $$y - 5z = -7 \implies y = -7 + 5z$$ 3. $$z$$ is a free variable. The solution can be written as: $$\{ (x, y, z) \,|\, x = -6 + 6z, \, y = -7 + 5z, \, z \in \mathbb{R} \}$$ --- ### Final Answer: **B. There are infinitely many solutions.** The solution is: $$\{ (x, y, z) \,|\, x = -6 + 6z, \, y = -7 + 5z, \, z \text{ is any real number} \}.$$ --- Would you like a detailed explanation of each step or clarification on specific parts? Here are some related questions to deepen your understanding: 1. What is the general form of an augmented matrix for systems of linear equations? 2. How do you identify free variables in a system of equations? 3. What does it mean for a system to be inconsistent? 4. What are the differences between Gaussian and Gauss-Jordan elimination? 5. How can matrix rank help determine the number of solutions to a system? **Tip:** Always check your row operations carefully to ensure the system remains equivalent!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Formulas
Row Operations
Gaussian Elimination
Theorems
Row Echelon Form
Inconsistent Systems
Suitable Grade Level
Grades 9-10
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