Find the solution of the following system using Gauss elimination. (Enter your answers as a comma-separated list.)

*x*

 − 

*y*

 − 

*z*

 + 

4*u*

 = 

-10

*x*

 − 

*y*

 + 

*z*

 + 

4*u*

 = 

-6

*y*

 + 

2*z*

 − 

*u*

 = 

8

4*y*

 + 

*z*

 + 

2*u*

 = 

12

(*x*, *y*, *z*, *u*) = ( 

We are given the following system of equations:

\[
\begin{align*}
x - y - z + 4u &= -10 \quad \text{(Equation 1)} \\
x - y + z + 4u &= -6 \quad \text{(Equation 2)} \\
y + 2z - u &= 8 \quad \text{(Equation 3)} \\
4y + z + 2u &= 12 \quad \text{(Equation 4)}
\end{align*}
\]

We will use Gaussian elimination to solve this system of equations. Let's write the augmented matrix for the system:

\[
\begin{pmatrix}
1 & -1 & -1 & 4 & -10 \\
1 & -1 & 1  & 4 & -6 \\
0 & 1  & 2  & -1 & 8 \\
0 & 4  & 1  & 2 & 12
\end{pmatrix}
\]

### Step 1: Eliminate \( x \)-terms from rows 2, 3, and 4

We subtract row 1 from row 2 to eliminate \( x \)-term in row 2:

\[
R_2 = R_2 - R_1 \implies \begin{pmatrix} 1 & -1 & -1 & 4 & -10 \\
0 & 0 & 2 & 0 & 4 \\
0 & 1 & 2 & -1 & 8 \\
0 & 4 & 1 & 2 & 12
\end{pmatrix}
\]

### Step 2: Eliminate \( y \)-terms from rows 3 and 4

Next, we subtract 4 times row 3 from row 4 to eliminate the \( y \)-term from row 4:

\[
R_4 = R_4 - 4R_3 \implies \begin{pmatrix} 1 & -1 & -1 & 4 & -10 \\
0 & 0 & 2 & 0 & 4 \\
0 & 1 & 2 & -1 & 8 \\
0 & 0 & -7 & 6 & -20
\end{pmatrix}
\]

### Step 3: Back-substitution

Now, we can start solving the system from the last row.

- From row 4: \(-7z + 6u = -20\)

Solving for \( z \):

\[
z = \frac{6u + 20}{7}
\]

We substitute this expression for \( z \) into row 3.

- From row 3: \( y + 2z - u = 8 \)

Substituting \( z = \frac{6u + 20}{7} \) into the equation:

\[
y + 2\left(\frac{6u + 20}{7}\right) - u = 8
\]

Simplifying, we solve for \( y \) and then substitute these into the other equations.

I'll compute the exact values for \( x, y, z, u \).The solution to the system of equations is:

\[
(x, y, z, u) = (-1, 3, 2, -1)
\]

Would you like further details on the steps, or do you have any questions?

Here are some related questions to consider:
1. How does Gaussian elimination work for non-square systems?
2. What is the significance of pivoting in Gaussian elimination?
3. Can Gaussian elimination always provide a unique solution?
4. How do row operations affect the determinant of a matrix?
5. What are some methods for solving systems of equations with more unknowns than equations?

**Tip:** Gaussian elimination is most efficient when dealing with systems that have a unique solution, but for systems with infinitely many solutions, it can be combined with parameterization techniques.

Find the solution of the following system using Gauss elimination. (Enter your answers as a comma-separated list.) x − y − z + 4*u = -10, x − y + z + 4*u = -6, y + 2*z − u = 8, 4*y + z + 2*u = 12. (x, y, z, u) = (

Learn how to solve a system of equations using Gaussian elimination. This example involves four variables (x, y, z, u) and demonstrates row operations and back-substitution. Suitable for students in Grades 10-12, this guide covers the key steps to solve the system efficiently.

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