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Solve the following linear systems by Gauss elimination, with partial pivoting if necessary (but without scaling). Show the intermediate steps. Check the result by substitution. If no solution or more than one solution exists, give a reason.

To solve this linear system using Gaussian elimination with partial pivoting, we will follow these steps:

1. **Set up the augmented matrix** for the system.
2. **Perform Gaussian elimination** to convert the matrix into an upper triangular form.
3. **Apply back-substitution** to find the solution for the variables.
4. **Check the result by substitution** into the original equations.

### Step 1: Set up the Augmented Matrix

Given system of equations:

\[
\begin{align*}
3.2x_1 + 1.6x_2 &= -0.8 \\
1.6x_1 - 0.8x_2 + 2.4x_3 &= 16.0 \\
2.4x_2 - 4.8x_3 + 3.6x_4 &= -39.0 \\
3.6x_3 + 2.4x_4 &= 10.2
\end{align*}
\]

The augmented matrix is:

\[
\begin{bmatrix}
3.2 & 1.6 & 0 & 0 & | & -0.8 \\
1.6 & -0.8 & 2.4 & 0 & | & 16.0 \\
0 & 2.4 & -4.8 & 3.6 & | & -39.0 \\
0 & 0 & 3.6 & 2.4 & | & 10.2
\end{bmatrix}
\]

### Step 2: Gaussian Elimination with Partial Pivoting

We will perform Gaussian elimination, making each leading coefficient of the rows below the pivot element zero.

#### a) First Pivot (Row 1)

Our pivot element is `3.2` in Row 1. No row interchange is needed.

- **Make element (2,1) zero**: 
  Multiply Row 1 by \( \frac{1.6}{3.2} = 0.5 \) and subtract from Row 2:
  
\[
R_2 = R_2 - 0.5 \cdot R_1
\]

- New Row 2:
\[
[1.6 - (0.5 \cdot 3.2), -0.8 - (0.5 \cdot 1.6), 2.4 - (0.5 \cdot 0), 0 - (0.5 \cdot 0) | 16 - (0.5 \cdot -0.8)]
\]

Simplify Row 2:
\[
R_2 = [0, -1.6, 2.4, 0 | 16.4]
\]

#### b) Second Pivot (Row 2)

Our next pivot element is `-1.6` in Row 2. No row interchange is needed.

- **Make element (3,2) zero**:
Multiply Row 2 by \( \frac{2.4}{-1.6} = -1.5 \) and add to Row 3:

\[
R_3 = R_3 + 1.5 \cdot R_2
\]

New Row 3:
\[
[0, 2.4 + (-1.5 \cdot -1.6), -4.8 + (-1.5 \cdot 2.4), 3.6 + (-1.5 \cdot 0) | -39 + (-1.5 \cdot 16.4)]
\]

Simplify Row 3:
\[
R_3 = [0, 0, -8.4, 3.6 | -63.6]
\]

#### c) Third Pivot (Row 3)

Our next pivot element is `-8.4` in Row 3. No row interchange is needed.

- **Make element (4,3) zero**:
Multiply Row 3 by \( \frac{3.6}{-8.4} \approx -0.4286 \) and add to Row 4:

\[
R_4 = R_4 + (-0.4286) \cdot R_3
\]

New Row 4:
\[
[0, 0, 3.6 + (-0.4286 \cdot -8.4), 2.4 + (-0.4286 \cdot 3.6) | 10.2 + (-0.4286 \cdot -63.6)]
\]

Simplify Row 4:
\[
R_4 = [0, 0, 0, 0 | -17.0]
\]

#### Observations After Elimination:

The last row simplifies to:
\[
0 = -17.0
\]

This is a contradiction, meaning that **no solution exists** for the given system of equations.

### Conclusion:

- Since the system leads to an inconsistency (a false equation like \(0 = -17.0\)), there is **no solution** for this linear system.

Would you like a detailed explanation of Gaussian elimination steps, or do you have any specific questions? 

### Related Questions:

1. What is partial pivoting in Gaussian elimination?
2. How do we handle a system with no unique solutions?
3. What are some methods for solving linear systems apart from Gaussian elimination?
4. Can Gaussian elimination be used for matrices of any size?
5. How does the elimination method differ when the matrix is sparse?

### Tip:

Always check for consistency in your linear system after performing Gaussian elimination. If you encounter an equation like \(0 = c\) (where \(c \neq 0\)), the system has no solution.

Learn how to solve linear systems using Gaussian elimination with partial pivoting. This example covers step-by-step transformation to upper triangular form, back-substitution, and consistency check. Ideal for undergraduate students studying linear algebra.

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