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

To solve the given system of equations using **Gauss Elimination**, we first express the system in augmented matrix form and then apply Gaussian elimination to transform the matrix into row-echelon form, followed by back substitution.

### Given system of equations:
\[
10x - 2y + 3z = 23 \tag{1}
\]
\[
2x + 10y - 5z = -33 \tag{2}
\]
\[
3x - 4y + 10z = 41 \tag{3}
\]

### Step 1: Write the augmented matrix
The augmented matrix corresponding to this system is:

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

### Step 2: Perform Gaussian Elimination
We will use row operations to reduce this matrix to upper triangular form.

#### Row 1: Normalize the first pivot by dividing row 1 by 10
\[
R_1 = \frac{1}{10} R_1 \quad \text{(Divide row 1 by 10)}
\]
\[
\begin{bmatrix}
1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\
2 & 10 & -5 & | & -33 \\
3 & -4 & 10 & | & 41
\end{bmatrix}
\]

#### Row 2: Eliminate the first element in row 2 by subtracting 2 times row 1 from row 2
\[
R_2 = R_2 - 2R_1
\]
\[
\begin{bmatrix}
1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\
0 & \frac{52}{5} & -\frac{31}{5} & | & -\frac{299}{5} \\
3 & -4 & 10 & | & 41
\end{bmatrix}
\]

#### Row 3: Eliminate the first element in row 3 by subtracting 3 times row 1 from row 3
\[
R_3 = R_3 - 3R_1
\]
\[
\begin{bmatrix}
1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\
0 & \frac{52}{5} & -\frac{31}{5} & | & -\frac{299}{5} \\
0 & -\frac{11}{5} & \frac{41}{10} & | & \frac{87}{10}
\end{bmatrix}
\]

### Step 3: Normalize the second pivot by multiplying row 2 by \( \frac{5}{52} \)
\[
R_2 = \frac{5}{52}R_2
\]
\[
\begin{bmatrix}
1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\
0 & 1 & -\frac{31}{52} & | & -\frac{299}{52} \\
0 & -\frac{11}{5} & \frac{41}{10} & | & \frac{87}{10}
\end{bmatrix}
\]

#### Row 3: Eliminate the second element in row 3 by adding \( \frac{11}{5} \) times row 2 to row 3
\[
R_3 = R_3 + \frac{11}{5}R_2
\]
\[
\begin{bmatrix}
1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\
0 & 1 & -\frac{31}{52} & | & -\frac{299}{52} \\
0 & 0 & \frac{579}{104} & | & \frac{2997}{104}
\end{bmatrix}
\]

### Step 4: Normalize the third pivot by multiplying row 3 by \( \frac{104}{579} \)
\[
R_3 = \frac{104}{579} R_3
\]
\[
\begin{bmatrix}
1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\
0 & 1 & -\frac{31}{52} & | & -\frac{299}{52} \\
0 & 0 & 1 & | & \frac{2997}{579}
\end{bmatrix}
\]

### Step 5: Perform back substitution
- From row 3: \( z = \frac{2997}{579} = 5.175 \)
- From row 2: \( y = -\frac{299}{52} + \frac{31}{52} \times 5.175 \)
- From row 1: \( x = \dots \)

I will now finalize these

Learn how to solve the system of equations 10x - 2y + 3z = 23, 2x + 10y - 5z = -33, 3x - 4y + 10z = 41 using the Gauss Elimination method. Follow this step-by-step tutorial with detailed matrix operations and back substitution.

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