Math Problem Statement
User responses cleared Homework:12.2 HW - Systems of Linear Equations: Matrices Question 30, 12.2.73 HW Score: 81.43%, 28.5 of 35 points Points: 0 of 1
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Question content area bottom Part 1 Select the correct choice below and, if necessary, fill in the answer box(es) in your choice. A. The solution is left parenthesis nothing comma 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 comma w right parenthesis | x equals nothing comma y equals nothing comma z equals nothing comma w is any real number EndSet . (Simplify your answers. Type expressions using w as the variable.) C. There are infinitely many solutions. The solution can be written as StartSet left parenthesis x comma y comma z comma w right parenthesis | x equals nothing comma y equals nothing comma z and w are any real numbers EndSet . (Simplify your answers. Type expressions using z and w as the variables.) D. There are infinitely many solutions. The solution can be written as StartSet left parenthesis x comma y comma z comma w right parenthesis | x equals nothing comma y comma z comma and w are any real numbers EndSet . (Simplify your answer. Type an expression using y, z, and w as the variables.) E. The system is inconsistent. input field 1input field 2input field 3input field 4input field 5input field 6input field 7input field 8input field 9input field 10Select the correct choice below and, if necessary, fill in the answer box(es) in your choice. Ask my instructor
Solution
To solve the system of equations using matrices and row operations, let's proceed step-by-step:
The given system:
We write the augmented matrix for the system:
6 & 1 & -1 & -1 & | & -5 \\ 1 & -1 & -3 & 2 & | & -4 \end{bmatrix}$$ ### Step 1: Normalize the first row Divide the first row by 6 to make the pivot in the first column 1: $$\begin{bmatrix} 1 & \frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & | & -\frac{5}{6} \\ 1 & -1 & -3 & 2 & | & -4 \end{bmatrix}$$ ### Step 2: Eliminate the first column for the second row Subtract the first row from the second row: $$\text{Row 2} \to \text{Row 2} - \text{Row 1}$$ $$\begin{bmatrix} 1 & \frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & | & -\frac{5}{6} \\ 0 & -\frac{7}{6} & -\frac{17}{6} & \frac{13}{6} & | & -\frac{19}{6} \end{bmatrix}$$ ### Step 3: Normalize the second row Multiply the second row by $$-6/7$$ to make the pivot in the second column 1: $$\begin{bmatrix} 1 & \frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & | & -\frac{5}{6} \\ 0 & 1 & \frac{17}{7} & -\frac{13}{7} & | & \frac{19}{7} \end{bmatrix}$$ ### Step 4: Eliminate the second column for the first row $$\text{Row 1} \to \text{Row 1} - \left(\frac{1}{6} \cdot \text{Row 2}\right)$$ Simplifying: $$\begin{bmatrix} 1 & 0 & -\frac{20}{42} & -\frac{10}{42} & | & -\frac{80}{42} \\ 0 & 1 & \frac{17}{7} & -\frac{13}{7} & | & \frac{19}{7} \end{bmatrix}$$ Simplify further: $$\begin{bmatrix} 1 & 0 & -\frac{10}{21} & -\frac{5}{21} & | & -\frac{40}{21} \\ 0 & 1 & \frac{17}{7} & -\frac{13}{7} & | & \frac{19}{7} \end{bmatrix}$$ ### Interpretation: The system is now in reduced row-echelon form. The solution depends on the variables $$z$$ and $$w$$, which are free variables. ### General solution: Let $$z = z$$ and $$w = w$$ (free variables). Then: $$x = -\frac{10}{21}z - \frac{5}{21}w - \frac{40}{21}$$ $$y = -\frac{17}{7}z + \frac{13}{7}w + \frac{19}{7}$$ Thus, the solution is **Option C**: $$\{(x, y, z, w) \ | \ x = -\frac{10}{21}z - \frac{5}{21}w - \frac{40}{21}, y = -\frac{17}{7}z + \frac{13}{7}w + \frac{19}{7}, z \text{ and } w \text{ are free variables}\}.$$ --- Do you need further clarification or detailed verification? Here are 5 related questions: 1. What is the process of reducing a matrix to row-echelon form? 2. How do you interpret free variables in solutions to systems of linear equations? 3. What does it mean if a system is inconsistent? 4. How would the solution change if the augmented matrix had a row of all zeros? 5. Can matrix row operations change the solution set of the system? **Tip:** Always check your row operations carefully to avoid computation errors.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Systems of Linear Equations
Matrix Row Operations
Gaussian Elimination
Formulas
Row operation rules: swapping rows, multiplying a row by a scalar, adding/subtracting rows
Reduced Row-Echelon Form (RREF)
Theorems
Existence and Uniqueness Theorem for Linear Systems
Suitable Grade Level
Grades 10-12 or early college
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