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

We are given the system of equations:

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

We'll solve this system using the **Gauss Elimination method**, which involves converting the system into an upper triangular form and then performing back substitution.

### Step 1: Write the system as an augmented matrix.
The system can be written in matrix form as:

\[
\begin{pmatrix}
10 & -2 & 3 & | & 23 \\
2 & 10 & -5 & | & -33 \\
3 & -4 & 10 & | & 41
\end{pmatrix}
\]

### Step 2: Perform row operations to obtain upper triangular form.

#### Operation 1: Eliminate the first variable (\(x\)) from the second and third rows.
- We will use row 1 to eliminate the \(x\)-terms in rows 2 and 3.

To eliminate \(x\) from the second row, we perform:
\[
R_2 \leftarrow R_2 - \frac{2}{10}R_1 = R_2 - 0.2R_1
\]

For the third row, we perform:
\[
R_3 \leftarrow R_3 - \frac{3}{10}R_1 = R_3 - 0.3R_1
\]

This gives the new matrix:

\[
\begin{pmatrix}
10 & -2 & 3 & | & 23 \\
0 & 10.4 & -5.6 & | & -37.6 \\
0 & -3.4 & 9.1 & | & 34.1
\end{pmatrix}
\]

#### Operation 2: Eliminate \(y\) from the third row.
We will use row 2 to eliminate the \(y\)-term from row 3. For this, perform:
\[
R_3 \leftarrow R_3 - \frac{-3.4}{10.4}R_2 = R_3 + 0.3269R_2
\]

This gives:

\[
\begin{pmatrix}
10 & -2 & 3 & | & 23 \\
0 & 10.4 & -5.6 & | & -37.6 \\
0 & 0 & 7.28 & | & 21.7738
\end{pmatrix}
\]

### Step 3: Back substitution.
We now have the system in upper triangular form, and we can perform back substitution.

#### From the third row:
\[
7.28z = 21.7738 \implies z = \frac{21.7738}{7.28} = 2.99 \approx 3
\]

#### From the second row:
\[
10.4y - 5.6z = -37.6
\]
Substituting \(z = 3\):
\[
10.4y - 5.6(3) = -37.6 \implies 10.4y - 16.8 = -37.6 \implies 10.4y = -20.8 \implies y = \frac{-20.8}{10.4} = -2
\]

#### From the first row:
\[
10x - 2y + 3z = 23
\]
Substituting \(y = -2\) and \(z = 3\):
\[
10x - 2(-2) + 3(3) = 23 \implies 10x + 4 + 9 = 23 \implies 10x + 13 = 23 \implies 10x = 10 \implies x = \frac{10}{10} = 1
\]

### Final solution:
\[
x = 1, \quad y = -2, \quad z = 3
\]

Would you like further details or explanations?

Here are 5 related questions to explore this topic further:
1. How does Gauss-Jordan elimination differ from Gauss elimination?
2. How do you determine when a system of equations has no solution using the Gauss elimination method?
3. What is the significance of obtaining an upper triangular matrix in this method?
4. How does the efficiency of Gaussian elimination compare to other methods like substitution or Cramer's rule?
5. Can Gaussian elimination be applied to non-square matrices?

**Tip**: Gaussian elimination can be easily adapted for solving systems of linear equations in larger dimensions, making it a versatile method in linear algebra.

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

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