Math Problem Statement

  1. Solve the system of equations by Gauss Elimination method 10x-2y+3z=23; 2x+10y-5z=-33; 3x-4y+10z=41. without decimal

Solution

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

  1. Write the system of equations in augmented matrix form.
  2. Perform row operations to get the matrix in row echelon form.
  3. Use back-substitution to find the values of xx, yy, and zz.

Step 1: Write the augmented matrix

The system of equations is:

10x - 2y + 3z &= 23, \\ 2x + 10y - 5z &= -33, \\ 3x - 4y + 10z &= 41. \end{aligned}$$ The corresponding augmented matrix is: $$\begin{pmatrix} 10 & -2 & 3 & | & 23 \\ 2 & 10 & -5 & | & -33 \\ 3 & -4 & 10 & | & 41 \end{pmatrix}$$ ### Step 2: Perform row operations **Row 1**: $$R_1 = (1/10)R_1$$ $$\begin{pmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 2 & 10 & -5 & | & -33 \\ 3 & -4 & 10 & | & 41 \end{pmatrix}$$ **Row 2**: $$R_2 = R_2 - 2R_1$$ $$\begin{pmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & \frac{52}{5} & -\frac{11}{5} & | & -\frac{386}{5} \\ 3 & -4 & 10 & | & 41 \end{pmatrix}$$ **Row 3**: $$R_3 = R_3 - 3R_1$$ $$\begin{pmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & \frac{52}{5} & -\frac{11}{5} & | & -\frac{386}{5} \\ 0 & -\frac{16}{5} & \frac{71}{10} & | & \frac{99}{10} \end{pmatrix}$$ **Row 2**: $$R_2 \times \textbf{5}$$ $$\begin{pmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & 52 & -11 & | & -386 \\ 0 & -16 & 71/10 & | & 99 \end{pmatrix}$$ **Continue for back-substitution!**

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Math Problem Analysis

Mathematical Concepts

Gauss Elimination Method
System of Linear Equations
Augmented Matrix

Formulas

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Theorems

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Suitable Grade Level

Advanced High School