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

Skip to Main content Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 Question 18 Question 19 Question 20 Question 21 Question 22 Question 23 Question 24 Question 25 Question 26 Question 27 Question 28 Question 29 Question 30 Question 31 Question 32 Question 33 Question 34 Question 35 Question content area top Part 1 Solve the following system of equations using matrices​ (row operations). If the system has no​ solution, say that it is inconsistent. left brace Start 2 By 3 Matrix 1st Row 1st Column 6 x plus y minus z minus w 2nd Column equals 3rd Column negative 5 2nd Row 1st Column x minus y minus 3 z plus 2 w 2nd Column equals 3rd Column negative 4 EndMatrix

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:

  1. 6x+yzw=56x + y - z - w = -5
  2. xy3z+2w=4x - y - 3z + 2w = -4

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

By Image

Drop file here or Click Here to upload

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