Math Problem Statement

Solve the system of linear equations using the Gauss-Jordan elimination method. 2x + 2y + z = 4 x + z = 2 4y − 3z = 12 (x, y, z) =

Solution

To solve this system of linear equations using the Gauss-Jordan elimination method, we start by representing the system in augmented matrix form:

2 & 2 & 1 & | & 4 \\ 1 & 0 & 1 & | & 2 \\ 0 & 4 & -3 & | & 12 \\ \end{bmatrix}$$ We aim to transform this matrix into reduced row echelon form (RREF), where each leading coefficient in each row is 1, and all other entries in the columns with leading coefficients are 0. ### Step-by-Step Solution 1. **Make the first entry of the first column a 1** by dividing the first row by 2: $$\begin{bmatrix} 1 & 1 & 0.5 & | & 2 \\ 1 & 0 & 1 & | & 2 \\ 0 & 4 & -3 & | & 12 \\ \end{bmatrix}$$ 2. **Eliminate the 1 in the first column of the second row** by subtracting the first row from the second row: $$\begin{bmatrix} 1 & 1 & 0.5 & | & 2 \\ 0 & -1 & 0.5 & | & 0 \\ 0 & 4 & -3 & | & 12 \\ \end{bmatrix}$$ 3. **Make the leading coefficient of the second row a 1** by multiplying the second row by -1: $$\begin{bmatrix} 1 & 1 & 0.5 & | & 2 \\ 0 & 1 & -0.5 & | & 0 \\ 0 & 4 & -3 & | & 12 \\ \end{bmatrix}$$ 4. **Eliminate the 1 in the second column of the first row** by subtracting the second row from the first row: $$\begin{bmatrix} 1 & 0 & 1 & | & 2 \\ 0 & 1 & -0.5 & | & 0 \\ 0 & 4 & -3 & | & 12 \\ \end{bmatrix}$$ 5. **Eliminate the 4 in the second column of the third row** by subtracting 4 times the second row from the third row: $$\begin{bmatrix} 1 & 0 & 1 & | & 2 \\ 0 & 1 & -0.5 & | & 0 \\ 0 & 0 & -1 & | & 12 \\ \end{bmatrix}$$ 6. **Make the leading coefficient of the third row a 1** by dividing the third row by -1: $$\begin{bmatrix} 1 & 0 & 1 & | & 2 \\ 0 & 1 & -0.5 & | & 0 \\ 0 & 0 & 1 & | & -12 \\ \end{bmatrix}$$ 7. **Eliminate the 1 in the third column of the first row** by subtracting the third row from the first row: $$\begin{bmatrix} 1 & 0 & 0 & | & 14 \\ 0 & 1 & -0.5 & | & 0 \\ 0 & 0 & 1 & | & -12 \\ \end{bmatrix}$$ 8. **Eliminate the -0.5 in the third column of the second row** by adding 0.5 times the third row to the second row: $$\begin{bmatrix} 1 & 0 & 0 & | & 14 \\ 0 & 1 & 0 & | & -6 \\ 0 & 0 & 1 & | & -12 \\ \end{bmatrix}$$ The matrix is now in reduced row echelon form, and we can read off the solutions for $$x$$, $$y$$, and $$z$$: $$x = 14, \quad y = -6, \quad z = -12$$ Thus, the solution is: $$(x, y, z) = (14, -6, -12)$$ Would you like a deeper explanation of any of these steps? --- ### Additional Related Questions: 1. How does Gauss-Jordan elimination differ from Gaussian elimination? 2. Can Gauss-Jordan elimination be applied to systems with no or infinite solutions? 3. What are pivot positions in the context of Gauss-Jordan elimination? 4. How can Gauss-Jordan elimination be implemented in matrix form? 5. What are some limitations of Gauss-Jordan elimination in solving large systems? #### Tip: To verify your solution, substitute the values of $$x$$, $$y$$, and $$z$$ back into the original equations to ensure they hold true.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Matrix Operations
Gauss-Jordan Elimination

Formulas

Row Reduction
Reduced Row Echelon Form (RREF)
Elementary Row Operations

Theorems

Gauss-Jordan Elimination Theorem

Suitable Grade Level

Grades 10-12

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