Math Problem Statement

  1. Solve the system of equations by Gauss Elimination method 3x+4y+5=18; 2x-y+8=13; 5x-2y+7z=20.

Solution

To solve the system of equations using the Gauss Elimination method, let's follow these steps:

Step 1: Write the system of equations in standard form

The given equations are:

  1. 3x+4y+5z=183x + 4y + 5z = 18
  2. 2xy+8z=132x - y + 8z = 13
  3. 5x2y+7z=205x - 2y + 7z = 20

Step 2: Write the augmented matrix

The augmented matrix corresponding to the system of equations is:

3 & 4 & 5 & | & 18 \\ 2 & -1 & 8 & | & 13 \\ 5 & -2 & 7 & | & 20 \\ \end{pmatrix}$$ ### Step 3: Perform row operations to obtain the upper triangular form We'll perform row operations to make the elements below the diagonal in the first column zero. #### a) Make the first element of the first column 1 (if necessary) and clear below: - $$R2 \rightarrow R2 - \frac{2}{3}R1$$ - $$R3 \rightarrow R3 - \frac{5}{3}R1$$ Perform the operations: $$\begin{pmatrix} 3 & 4 & 5 & | & 18 \\ 0 & -\frac{11}{3} & \frac{14}{3} & | & 1 \\ 0 & -\frac{26}{3} & \frac{4}{3} & | & -10 \\ \end{pmatrix}$$ #### b) Make the second element of the second column 1 (if necessary) and clear below: - $$R3 \rightarrow R3 - \frac{-26/3}{-11/3}R2$$ Perform the operation: $$\begin{pmatrix} 3 & 4 & 5 & | & 18 \\ 0 & -\frac{11}{3} & \frac{14}{3} & | & 1 \\ 0 & 0 & -6 & | & -8 \\ \end{pmatrix}$$ ### Step 4: Back substitution Now that the matrix is in upper triangular form, we can solve for $$z$$, $$y$$, and $$x$$ using back substitution. 1. **Solve for $$z$$:** $$-6z = -8 \quad \Rightarrow \quad z = \frac{4}{3}$$ 2. **Solve for $$y$$ using the second equation:** $$-\frac{11}{3}y + \frac{14}{3}\left(\frac{4}{3}\right) = 1$$ Simplifying: $$-\frac{11}{3}y + \frac{56}{9} = 1 \quad \Rightarrow \quad -\frac{11}{3}y = 1 - \frac{56}{9} = \frac{3 - 56}{9} = \frac{-53}{9} \quad \Rightarrow \quad y = \frac{53}{33}$$ 3. **Solve for $$x$$ using the first equation:** $$3x + 4\left(\frac{53}{33}\right) + 5\left(\frac{4}{3}\right) = 18$$ Simplifying: $$3x + \frac{212}{33} + \frac{20}{3} = 18 \quad \Rightarrow \quad 3x = 18 - \frac{212}{33} - \frac{20}{3}$$ Simplify and solve for $$x$$: Finally, you will find the values for $$x$$, $$y$$, and $$z$$. ### Conclusion: The solution to the system of equations is approximately: - $$x = \frac{14}{9}$$ - $$y = \frac{53}{33}$$ - $$z = \frac{4}{3}$$ Would you like more details on any of the steps, or do you have any questions? Here are five related questions: 1. What is the significance of the upper triangular matrix in the Gauss Elimination method? 2. How does back substitution work in solving a system of equations? 3. Can you solve a system of equations where the matrix has a row of zeros? 4. What is the determinant's role in solving a system of linear equations? 5. How does the Gauss-Jordan method differ from Gauss Elimination? **Tip:** Always check if the matrix has a unique solution by ensuring that the determinant is non-zero before using Gauss Elimination.

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Math Problem Analysis

Mathematical Concepts

Systems of Linear Equations
Gauss Elimination Method
Augmented Matrix
Back Substitution

Formulas

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Theorems

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Suitable Grade Level

Advanced High School