2. Solve the system of equations by Gauss Elimination method 10x-2y+3=23; 2x+10y-5=-33; 3x-4y+10z=41. 

To solve the system of equations using the Gauss Elimination method, we first need to write the equations in standard form and then represent them as an augmented matrix. The given system of equations is:

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

### Step 1: Convert the equations into standard form.
The equations should be in the form \(ax + by + cz = d\):

1. \(10x - 2y + 0z = 23 - 3 \Rightarrow 10x - 2y = 20\)
2. \(2x + 10y + 0z = -33 + 5 \Rightarrow 2x + 10y = -28\)
3. \(3x - 4y + 10z = 41\)

This results in the following system:

\[
\begin{aligned}
1. & \quad 10x - 2y + 0z = 20 \\
2. & \quad 2x + 10y + 0z = -28 \\
3. & \quad 3x - 4y + 10z = 41 
\end{aligned}
\]

### Step 2: Represent the system as an augmented matrix.
\[
\begin{pmatrix}
10 & -2 & 0 & | & 20 \\
2 & 10 & 0 & | & -28 \\
3 & -4 & 10 & | & 41
\end{pmatrix}
\]

### Step 3: Apply Gauss Elimination to reduce the matrix to row-echelon form.

We start with the first row (pivoting on 10 in the first column) and eliminate the first column below the pivot.

1. **Row 2 Operation**: Subtract \(\frac{1}{5}\) of Row 1 from Row 2:
   \[
   R_2 = R_2 - \frac{1}{5}R_1
   \]
2. **Row 3 Operation**: Subtract \(\frac{3}{10}\) of Row 1 from Row 3:
   \[
   R_3 = R_3 - \frac{3}{10}R_1
   \]

This results in:

\[
\begin{pmatrix}
10 & -2 & 0 & | & 20 \\
0 & 10.4 & 0 & | & -32 \\
0 & -2.4 & 10 & | & 35
\end{pmatrix}
\]

Next, eliminate the second column below the pivot at \(R_2C_2\) using Row 2:

3. **Row 3 Operation**: Add \(\frac{12}{52}\) of Row 2 to Row 3:
   \[
   R_3 = R_3 + \frac{6}{13}R_2
   \]

This results in:

\[
\begin{pmatrix}
10 & -2 & 0 & | & 20 \\
0 & 10.4 & 0 & | & -32 \\
0 & 0 & 10 & | & 12
\end{pmatrix}
\]

### Step 4: Perform back substitution.
Use the triangular form to solve.

1. From Row 3:
   \[
   z = \frac{12}{10} = 1.2
   \]
2. From Row 2:
   \[
   10.4y = -32 \quad \Rightarrow y = \frac{-32}{10.4} \approx -3.077
   \]
3. From Row 1:
   \[
   10x - 2(-3.077) = 20 \quad \Rightarrow x \approx 1.385
   \]

Thus, the solution is:

\[
\boxed{x = 1.385, \, y = -3.077, \, z = 1.2}
\]

Would you like to explore this further, or is there another question I can assist you with?

### Related Questions:
1. How does Gauss-Jordan elimination differ from Gauss elimination?
2. Can all systems be solved using Gaussian elimination?
3. What are the advantages of using matrix methods in solving systems of equations?
4. How do you determine if a system has no solution or infinite solutions using elimination?
5. How is Gaussian elimination applied in larger systems?

### Tip:
Always ensure the leading coefficient (pivot) is non-zero when performing row operations in Gauss elimination to avoid errors in computation.

Learn how to solve a system of equations using the Gauss Elimination method step-by-step. This problem involves linear algebra concepts, including augmented matrix representation. Suitable for college-level students.

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