Math Problem Statement
Solve using Gauss-Jordan elimination. 2x 1plus 2x 2minus 5x 3equals negative 1 4x 1plus 16x 2minus 38x 3equals 54 x 1plus 3x 2minus 7x 3equals 9 Question content area bottom Part 1 Select the correct choice below and fill in the answer box(es) within your choice. A. The unique solution is x 1equals negative 7,x 2equals 4, and x 3equals 2. B. The system has infinitely many solutions. The solution is x 1equals enter your response here, x 2equals enter your response here, and x 3equalst. (Simplify your answers. Type expressions using t as the variable.) C. The system has infinitely many solutions. The solution is x 1equals enter your response here, x 2equalss, and x 3equalst. (Simplify your answer. Type an expression using s and t as the variables.) D. There is no solution.
Solution
To solve the given system of equations using Gauss-Jordan elimination, we start by writing the augmented matrix for the system.
Step 1: Write the augmented matrix
The system of equations is:
The augmented matrix corresponding to this system is:
2 & 2 & -5 & | & -1 \\ 4 & 16 & -38 & | & 54 \\ 1 & 3 & -7 & | & 9 \end{pmatrix}$$ ### Step 2: Perform row operations to get the matrix in reduced row echelon form (RREF). #### **First, make the leading entry in the first column equal to 1:** Divide Row 1 by 2: $$R_1 \rightarrow \frac{R_1}{2}$$ $$\begin{pmatrix} 1 & 1 & -\frac{5}{2} & | & -\frac{1}{2} \\ 4 & 16 & -38 & | & 54 \\ 1 & 3 & -7 & | & 9 \end{pmatrix}$$ #### **Next, eliminate the first column's entries in rows 2 and 3:** - For Row 2: Subtract $$4 \times R_1$$ from Row 2: $$R_2 \rightarrow R_2 - 4 \times R_1$$ - For Row 3: Subtract $$1 \times R_1$$ from Row 3: $$R_3 \rightarrow R_3 - R_1$$ Performing these operations: $$\begin{pmatrix} 1 & 1 & -\frac{5}{2} & | & -\frac{1}{2} \\ 0 & 12 & -\frac{18}{2} & | & 56 \\ 0 & 2 & -\frac{9}{2} & | & \frac{19}{2} \end{pmatrix}$$ #### **Now, normalize Row 2 to make the leading coefficient equal to 1:** Divide Row 2 by 12: $$R_2 \rightarrow \frac{R_2}{12}$$ $$\begin{pmatrix} 1 & 1 & -\frac{5}{2} & | & -\frac{1}{2} \\ 0 & 1 & -\frac{3}{2} & | & \frac{14}{3} \\ 0 & 2 & -\frac{9}{2} & | & \frac{19}{2} \end{pmatrix}$$ #### **Eliminate the second column's entries in Rows 1 and 3:** - For Row 1: Subtract $$R_2$$ from Row 1: $$R_1 \rightarrow R_1 - R_2$$ - For Row 3: Subtract $$2 \times R_2$$ from Row 3: $$R_3 \rightarrow R_3 - 2 \times R_2$$ After performing these operations: $$\begin{pmatrix} 1 & 0 & -1 & | & -\frac{25}{3} \\ 0 & 1 & -\frac{3}{2} & | & \frac{14}{3} \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$$ ### Step 3: Interpret the result The resulting matrix corresponds to the system: $$x_1 - x_3 = -\frac{25}{3}$$ $$x_2 - \frac{3}{2}x_3 = \frac{14}{3}$$ $$0 = 0$$ From the third equation, we have no new information (it's always true), so we have **infinitely many solutions** depending on $$x_3$$. Let $$x_3 = t$$ (where $$t$$ is a free variable). From the first equation: $$x_1 = -\frac{25}{3} + t$$ From the second equation: $$x_2 = \frac{14}{3} + \frac{3}{2}t$$ ### Step 4: Conclusion The system has infinitely many solutions, and the solution can be expressed as: $$x_1 = -\frac{25}{3} + t, \quad x_2 = \frac{14}{3} + \frac{3}{2}t, \quad x_3 = t$$ where $$t$$ is any real number. Thus, the correct choice is **B**: The system has infinitely many solutions. The solution is: $$x_1 = -\frac{25}{3} + t, \quad x_2 = \frac{14}{3} + \frac{3}{2}t, \quad x_3 = t$$ Let me know if you'd like more details or have any further questions! ### Additional Related Questions: 1. How do you perform Gaussian elimination, and how does it differ from Gauss-Jordan elimination? 2. What is the geometric interpretation of the solution to a system of linear equations? 3. How do you check if a system of equations has no solution using row reduction? 4. What conditions must be met for a system to have a unique solution? 5. How can you solve a system of equations using matrix inverses? ### Tip: In Gaussian elimination, always aim to get a row of zeros in the last row if there is no contradiction in the system (like 0 = 1). If you see that, it means no solution exists.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gauss-Jordan Elimination
Systems of Linear Equations
Matrices
Formulas
Gauss-Jordan elimination
Augmented matrix
Theorems
Gauss-Jordan elimination theorem
Suitable Grade Level
Grades 10-12