Math Problem Statement

Use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If the system is dependent, set z = a and solve for x and y in terms of a.)

2x − y + 3z = 14 2y − z = 19 7x − 5y = 1 (x, y, z) =

, ,

Solution

We are given the following system of equations:

1. & \quad 2x - y + 3z = 14 \\ 2. & \quad 2y - z = 19 \\ 3. & \quad 7x - 5y = 1 \end{aligned}$$ We will use Gaussian elimination or Gauss-Jordan elimination to solve this system of equations by transforming the augmented matrix into row-echelon form, and then applying back-substitution. ### Step 1: Write the augmented matrix We first convert the system into an augmented matrix form. $$\begin{pmatrix} 2 & -1 & 3 & | & 14 \\ 0 & 2 & -1 & | & 19 \\ 7 & -5 & 0 & | & 1 \end{pmatrix}$$ ### Step 2: Apply Gaussian elimination Our goal is to eliminate the terms below the pivot in the first column. We will perform row operations to zero out the elements below the first pivot (the element in row 1, column 1). **1. Eliminate the element in row 3, column 1:** We perform the row operation $$R_3 \rightarrow R_3 - \frac{7}{2}R_1$$. $$\text{New row 3} = (7, -5, 0 | 1) - \frac{7}{2}(2, -1, 3 | 14) = (0, -\frac{3}{2}, -\frac{21}{2} | -48)$$ The updated augmented matrix becomes: $$\begin{pmatrix} 2 & -1 & 3 & | & 14 \\ 0 & 2 & -1 & | & 19 \\ 0 & -\frac{3}{2} & -\frac{21}{2} & | & -48 \end{pmatrix}$$ ### Step 3: Simplify row 3 To make calculations easier, let's multiply row 3 by 2 to eliminate fractions: $$R_3 \rightarrow 2R_3$$ This gives us: $$\begin{pmatrix} 2 & -1 & 3 & | & 14 \\ 0 & 2 & -1 & | & 19 \\ 0 & -3 & -21 & | & -96 \end{pmatrix}$$ ### Step 4: Eliminate the element in row 3, column 2 We now perform the row operation $$R_3 \rightarrow R_3 + \frac{3}{2}R_2$$ to eliminate the element in row 3, column 2. $$R_3 = (0, -3, -21 | -96) + \frac{3}{2}(0, 2, -1 | 19) = (0, 0, -22.5 | -67.5)$$ Thus, the updated matrix is: $$\begin{pmatrix} 2 & -1 & 3 & | & 14 \\ 0 & 2 & -1 & | & 19 \\ 0 & 0 & -22.5 & | & -67.5 \end{pmatrix}$$ ### Step 5: Solve using back-substitution We can now solve for $$z$$, $$y$$, and $$x$$. **From the third row:** $$-22.5z = -67.5 \quad \Rightarrow \quad z = 3$$ **Substitute $$z = 3$$ into the second row:** $$2y - z = 19 \quad \Rightarrow \quad 2y - 3 = 19 \quad \Rightarrow \quad 2y = 22 \quad \Rightarrow \quad y = 11$$ **Substitute $$z = 3$$ and $$y = 11$$ into the first row:** $$2x - y + 3z = 14 \quad \Rightarrow \quad 2x - 11 + 9 = 14 \quad \Rightarrow \quad 2x - 2 = 14 \quad \Rightarrow \quad 2x = 16 \quad \Rightarrow \quad x = 8$$ ### Final Solution: The solution to the system of equations is: $$(x, y, z) = (8, 11, 3)$$ Would you like a more detailed explanation, or do you have any questions? Here are 5 related questions: 1. How does Gaussian elimination differ from Gauss-Jordan elimination? 2. Can a system of equations have more than one solution? How would you identify that case? 3. How can you determine if a system of equations is inconsistent? 4. What does it mean for a system of equations to be dependent? 5. How do you represent a system of equations in matrix form? **Tip:** When solving a system of linear equations using matrices, always check the determinant of the coefficient matrix to quickly assess if the system is invertible.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
System of Linear Equations
Matrices
Gaussian Elimination
Gauss-Jordan Elimination

Formulas

Row Operations for Matrix Transformation
Back-Substitution in Row-Echelon Form
Gauss-Jordan Elimination to Reduced Row-Echelon Form

Theorems

Gaussian Elimination Process
Gauss-Jordan Elimination Method

Suitable Grade Level

Grades 10-12

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