Math Problem Statement

Solve the following system of equations 3 π‘₯ + 𝑦 βˆ’ 𝑧 = 3, 2π‘₯ βˆ’ 8𝑦 + 𝑧 = βˆ’5, π‘₯ βˆ’ 2𝑦 + 9𝑧 = 8 by Gauss elimination metho

Solution

To solve the given system of equations using the Gauss elimination method, we proceed as follows:

System of Equations:

  1. 3x+yβˆ’z=33x + y - z = 3
  2. 2xβˆ’8y+z=βˆ’52x - 8y + z = -5
  3. xβˆ’2y+9z=8x - 2y + 9z = 8

Step 1: Represent the system in augmented matrix form

The augmented matrix of the system is:

3 & 1 & -1 & 3 \\ 2 & -8 & 1 & -5 \\ 1 & -2 & 9 & 8 \end{bmatrix}$$ ### Step 2: Perform row operations to reduce the matrix to row-echelon form. #### (a) Make the pivot in the first column $$1$$ (using row 3): $$R_1 \leftrightarrow R_3$$ $$\begin{bmatrix} 1 & -2 & 9 & 8 \\ 2 & -8 & 1 & -5 \\ 3 & 1 & -1 & 3 \end{bmatrix}$$ #### (b) Eliminate the entries below the pivot in column 1: $$R_2 \to R_2 - 2R_1, \quad R_3 \to R_3 - 3R_1$$ $$\begin{bmatrix} 1 & -2 & 9 & 8 \\ 0 & -4 & -17 & -21 \\ 0 & 7 & -28 & -21 \end{bmatrix}$$ #### (c) Make the pivot in the second column $$1$$ (dividing $$R_2$$ by $$-4$$): $$R_2 \to \frac{R_2}{-4}$$ $$\begin{bmatrix} 1 & -2 & 9 & 8 \\ 0 & 1 & \frac{17}{4} & \frac{21}{4} \\ 0 & 7 & -28 & -21 \end{bmatrix}$$ #### (d) Eliminate the entry below the pivot in column 2: $$R_3 \to R_3 - 7R_2$$ $$\begin{bmatrix} 1 & -2 & 9 & 8 \\ 0 & 1 & \frac{17}{4} & \frac{21}{4} \\ 0 & 0 & -\frac{180}{4} & -\frac{192}{4} \end{bmatrix}$$ $$\begin{bmatrix} 1 & -2 & 9 & 8 \\ 0 & 1 & \frac{17}{4} & \frac{21}{4} \\ 0 & 0 & -45 & -48 \end{bmatrix}$$ #### (e) Make the pivot in the third column $$1$$: $$R_3 \to \frac{R_3}{-45}$$ $$\begin{bmatrix} 1 & -2 & 9 & 8 \\ 0 & 1 & \frac{17}{4} & \frac{21}{4} \\ 0 & 0 & 1 & \frac{16}{15} \end{bmatrix}$$ #### (f) Eliminate the entries above the pivot in column 3: 1. $$R_2 \to R_2 - \frac{17}{4}R_3$$ 2. $$R_1 \to R_1 - 9R_3$$ $$\begin{bmatrix} 1 & -2 & 0 & \frac{28}{15} \\ 0 & 1 & 0 & \frac{19}{15} \\ 0 & 0 & 1 & \frac{16}{15} \end{bmatrix}$$ #### (g) Eliminate the entry above the pivot in column 2: $$R_1 \to R_1 + 2R_2$$ $$\begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & \frac{19}{15} \\ 0 & 0 & 1 & \frac{16}{15} \end{bmatrix}$$ ### Step 3: Extract the solution: From the matrix: $$x = 2, \quad y = \frac{19}{15}, \quad z = \frac{16}{15}$$ ### Final Answer: $$x = 2, \quad y = \frac{19}{15}, \quad z = \frac{16}{15}$$ --- Would you like further details or help verifying these steps? 😊 ### Related Questions: 1. What is the difference between Gauss elimination and Gauss-Jordan elimination? 2. How do you verify the solution of a system of equations? 3. What are the conditions for a unique solution in a system of linear equations? 4. How does matrix row reduction help in solving equations? 5. What if the determinant of the coefficient matrix is zero? ### Tip: Always check for consistency (dependent or inconsistent systems) when solving systems of equations!

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Gaussian Elimination

Formulas

Row operations to achieve row-echelon form
Back substitution for solution extraction

Theorems

Gaussian Elimination Method

Suitable Grade Level

Grades 11-12 and Undergraduate