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

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 \(x\), \(y\), and \(z\).

### Step 1: Write the augmented matrix

The system of equations is:

\[
\begin{aligned}
10x - 2y + 3z &= 23, \\
2x + 10y - 5z &= -33, \\
3x - 4y + 10z &= 41.
\end{aligned}
\]

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!**


Learn how to solve a system of equations using the Gauss Elimination method. Detailed step-by-step solution for 10x - 2y + 3z = 23, 2x + 10y - 5z = -33, 3x - 4y + 10z = 41. Suitable for advanced high school students.

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