Math Problem Statement
y+z-2w=0 2x-3y-3z+6w=2 4x+y+z=4 Solve using guass method
Solution
To solve the system of equations using the Gaussian elimination method, we first write the system as an augmented matrix:
0 & 1 & 1 & -2 & | & 0 \\ 2 & -3 & -3 & 6 & | & 2 \\ 4 & 1 & 1 & 0 & | & 4 \end{bmatrix}$$ Now, we will perform row operations to get the matrix in row echelon form and eventually solve for the variables. ### Step 1: Swap Row 1 and Row 2 Since we want the first element of the first row to be non-zero, we swap row 1 and row 2. $$\begin{bmatrix} 2 & -3 & -3 & 6 & | & 2 \\ 0 & 1 & 1 & -2 & | & 0 \\ 4 & 1 & 1 & 0 & | & 4 \end{bmatrix}$$ ### Step 2: Make the first column below the pivot element zero Now, we need to eliminate the 4 in the third row, first column. We can do this by performing $$R_3 = R_3 - 2R_1$$. $$R_3 = R_3 - 2R_1 \rightarrow \begin{bmatrix} 2 & -3 & -3 & 6 & | & 2 \\ 0 & 1 & 1 & -2 & | & 0 \\ 0 & 7 & 7 & -12 & | & 0 \end{bmatrix}$$ ### Step 3: Make the second column below the pivot zero Next, we need to eliminate the 7 in the third row, second column. We can do this by performing $$R_3 = R_3 - 7R_2$$. $$R_3 = R_3 - 7R_2 \rightarrow \begin{bmatrix} 2 & -3 & -3 & 6 & | & 2 \\ 0 & 1 & 1 & -2 & | & 0 \\ 0 & 0 & 0 & 2 & | & 0 \end{bmatrix}$$ ### Step 4: Make the third column a pivot (if needed) We see that the third column in the third row is all zeros except for the right-hand side, which is already 0. This indicates that the system has infinitely many solutions and is dependent. The last equation simplifies to: $$0 = 0$$ ### Step 5: Back Substitution Now, we perform back substitution to express the variables in terms of free variables. From the augmented matrix: 1. $$2x - 3y - 3z + 6w = 2$$ 2. $$y + z - 2w = 0$$ 3. $$0 = 0$$ From equation (2): $$y = -z + 2w$$ Substitute this expression for $$y$$ into equation (1): $$2x - 3(-z + 2w) - 3z + 6w = 2$$ Simplify: $$2x + 3z - 6w - 3z + 6w = 2$$ $$2x = 2$$ $$x = 1$$ Now, we know $$x = 1$$, and from equation (2), $$y = -z + 2w$$. The system is underdetermined, so $$z$$ and $$w$$ are free variables. ### Final Solution: $$x = 1, \quad y = -z + 2w, \quad z = z, \quad w = w$$ where $$z$$ and $$w$$ are free variables. --- Would you like to explore this solution further or have any other questions? Here are some related questions that may interest you: 1. What are the geometric interpretations of the solution (in terms of planes)? 2. How would you use Gaussian elimination for a system with no solutions? 3. What is the role of the rank of a matrix in determining the number of solutions? 4. How do you handle inconsistent systems using Gaussian elimination? 5. How can Gaussian elimination be used for solving larger systems of equations? Tip: When solving systems with infinitely many solutions, always express the free variables and ensure that your solution represents the entire solution space.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Formulas
Augmented matrix representation of a system of linear equations
Row operations: row addition, scalar multiplication, row swapping
Theorems
Gaussian Elimination Method
Row Echelon Form
Back Substitution
Suitable Grade Level
Grades 10-12
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