Math Problem Statement

Find the general solutions of the systems whose augmented matrices are given in Exercises 7

|1 -7 0 6 5| |0 0 1 -2 -3| |-1 7 -4 2 7|

Solution

To solve the system of linear equations corresponding to the augmented matrix, we'll first write out the matrix, perform row operations to convert it into its reduced row echelon form (RREF), and then interpret the solutions.

Step 1: Write the augmented matrix

The given augmented matrix is:

1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ -1 & 7 & -4 & 2 & 7 \end{pmatrix}$$ This corresponds to the system of linear equations: $$\begin{aligned} 1x_1 & - 7x_2 + 0x_3 + 6x_4 &= 5 \quad & \text{(Equation 1)} \\ 0x_1 & + 0x_2 + 1x_3 - 2x_4 &= -3 \quad & \text{(Equation 2)} \\ -1x_1 & + 7x_2 - 4x_3 + 2x_4 &= 7 \quad & \text{(Equation 3)} \end{aligned}$$ ### Step 2: Perform row operations to reach RREF 1. **Row 3 Operation:** Add Row 1 to Row 3 to eliminate $$x_1$$ in Row 3. $$R_3 \to R_3 + R_1$$ The new matrix becomes: $$\begin{pmatrix} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & -4 & 8 & 12 \end{pmatrix}$$ 2. **Row 3 Operation:** Divide Row 3 by $$-4$$ to simplify it. $$R_3 \to \frac{1}{-4}R_3$$ The new matrix becomes: $$\begin{pmatrix} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & 1 & -2 & -3 \end{pmatrix}$$ 3. **Row 3 Operation:** Subtract Row 2 from Row 3 to eliminate $$x_3$$ in Row 3. $$R_3 \to R_3 - R_2$$ The new matrix becomes: $$\begin{pmatrix} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$ Now the matrix is in row echelon form (not quite reduced yet, but it's sufficient to find the general solution). ### Step 3: Interpret the solutions The matrix represents the following system: $$\begin{aligned} x_1 & - 7x_2 + 6x_4 &= 5 \quad & \text{(Equation 1)} \\ x_3 & - 2x_4 &= -3 \quad & \text{(Equation 2)} \\ 0 &= 0 \quad & \text{(Equation 3)} \end{aligned}$$ - $$x_2$$ and $$x_4$$ are free variables (can take any value). - Express $$x_1$$ and $$x_3$$ in terms of the free variables $$x_2$$ and $$x_4$$. ### General Solution: Let $$x_2 = t$$ and $$x_4 = s$$, where $$t$$ and $$s$$ are any real numbers. 1. From Equation 2: $$x_3 = 2s - 3$$ 2. From Equation 1: $$x_1 = 7t - 6s + 5$$ Thus, the general solution is: $$\begin{aligned} x_1 &= 7t - 6s + 5 \\ x_2 &= t \\ x_3 &= 2s - 3 \\ x_4 &= s \end{aligned}$$ where $$t$$ and $$s$$ are free parameters. ### Do you want further details or have any questions? --- Here are 5 related questions you might find interesting: 1. How do you identify free variables in a system of linear equations? 2. What is the difference between row echelon form (REF) and reduced row echelon form (RREF)? 3. How do you determine if a system of linear equations has a unique solution, infinitely many solutions, or no solution? 4. How does the concept of linear independence relate to solving systems of equations? 5. How can you verify the correctness of the general solution found for a system of linear equations? **Tip:** Always check your final solution by substituting it back into the original system of equations. This ensures accuracy.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Matrix Operations

Formulas

Row Operations
Reduced Row Echelon Form (RREF)

Theorems

Solution of a Linear System
Free and Basic Variables

Suitable Grade Level

Undergraduate Level or Advanced High School (Grades 11-12)

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