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

To solve the system of equations using the Gauss Elimination method, we first write the system in augmented matrix form and then perform row operations to convert it into an upper triangular matrix. Finally, we'll perform back-substitution to find the solution.

### Given system of equations:
\[
\begin{align*}
10x - 2y + 3z &= 23 \quad \text{(Equation 1)} \\
2x + 10y - 5z &= -33 \quad \text{(Equation 2)} \\
3x - 4y + 10z &= 41 \quad \text{(Equation 3)}
\end{align*}
\]

### Step 1: Write the augmented matrix
\[
\begin{pmatrix}
10 & -2 & 3 & \vert & 23 \\
2 & 10 & -5 & \vert & -33 \\
3 & -4 & 10 & \vert & 41
\end{pmatrix}
\]

### Step 2: Perform row operations to form an upper triangular matrix

**Row 1 (R1)** is already fine, so we'll use it to eliminate the \(x\) terms from Row 2 and Row 3.

**Operation 1:**
Eliminate the \(x\)-term from Row 2 using \( R2 \) and \( R1 \):
\[
R2 \rightarrow R2 - \frac{2}{10}R1 = R2 - \frac{1}{5}R1
\]

\[
\begin{pmatrix}
10 & -2 & 3 & \vert & 23 \\
0 & \left(10 + \frac{2}{5} \times 2 \right) & \left(-5 - \frac{3}{5} \right) & \vert & \left(-33 - \frac{23}{5}\right) \\
3 & -4 & 10 & \vert & 41
\end{pmatrix}
\]

Simplifying the second row:
\[
\begin{pmatrix}
10 & -2 & 3 & \vert & 23 \\
0 & 10.4 & -5.6 & \vert & -37.6 \\
3 & -4 & 10 & \vert & 41
\end{pmatrix}
\]

**Operation 2:**
Eliminate the \(x\)-term from Row 3 using \( R3 \) and \( R1 \):
\[
R3 \rightarrow R3 - \frac{3}{10}R1
\]
\[
\begin{pmatrix}
10 & -2 & 3 & \vert & 23 \\
0 & 10.4 & -5.6 & \vert & -37.6 \\
0 & -3.4 & 9.1 & \vert & 34.1
\end{pmatrix}
\]

### Step 3: Make the matrix upper triangular
Next, we'll make the lower left element of the 3rd row zero by eliminating the \(y\) term in the 3rd row using \( R3 \) and \( R2 \):

**Operation 3:**
Eliminate the \(y\)-term from Row 3:
\[
R3 \rightarrow R3 + \frac{3.4}{10.4}R2 = R3 + \frac{17}{52}R2
\]
\[
\begin{pmatrix}
10 & -2 & 3 & \vert & 23 \\
0 & 10.4 & -5.6 & \vert & -37.6 \\
0 & 0 & 6.4 & \vert & 11.2
\end{pmatrix}
\]

### Step 4: Back-substitution

Now that the matrix is upper triangular, we can solve for \(z\), \(y\), and \(x\) by back-substitution:

From \( R3 \):
\[
6.4z = 11.2 \quad \Rightarrow \quad z = \frac{11.2}{6.4} = 1.75
\]

From \( R2 \):
\[
10.4y - 5.6z = -37.6 \quad \Rightarrow \quad 10.4y - 5.6(1.75) = -37.6
\]
\[
10.4y - 9.8 = -37.6 \quad \Rightarrow \quad 10.4y = -27.8 \quad \Rightarrow \quad y = \frac{-27.8}{10.4} = -2.67
\]

From \( R1 \):
\[
10x - 2y + 3z = 23 \quad \Rightarrow \quad 10x - 2(-2.67) + 3(1.75) = 23
\]
\[
10x + 5.34 + 5.25 = 23 \quad \Rightarrow \quad 10x = 12.41 \quad \Rightarrow \quad x = 1.241
\]

### Solution
\[
x = 1.241, \quad y = -2.67, \quad z = 1.75
\]

These are the solutions to the given system of equations.

Would you like further details or have any questions about the method used?

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**Relative Questions:**
1. How can we check the solution of a system of equations solved by Gauss Elimination?
2. What are the advantages of using Gauss Elimination over other methods like substitution or graphing?
3. How does the Gauss-Jordan method differ from the Gauss Elimination method?
4. What are the potential issues with Gauss Elimination when dealing with large matrices?
5. How can we apply Gauss Elimination to non-square systems of equations?

**Tip:** Always verify your solutions by substituting them back into the original equations to ensure they satisfy all the given equations.

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, including matrix operations and back-substitution. Suitable for undergraduate level students.

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