2x-3y-5z=1, 3x+y-2z=-4, x-2y+z=5 by Gaussian method

To solve the system of linear equations using the Gaussian elimination method, we follow these steps:

Given system of equations:
1. \( 2x - 3y - 5z = 1 \)
2. \( 3x + y - 2z = -4 \)
3. \( x - 2y + z = 5 \)

**Step 1: Write the system as an augmented matrix.**

\[
\begin{pmatrix}
2 & -3 & -5 & | & 1 \\
3 & 1 & -2 & | & -4 \\
1 & -2 & 1 & | & 5 \\
\end{pmatrix}
\]

**Step 2: Perform row operations to get a triangular (upper triangular) form.**

We begin by making the coefficient of \(x\) in the first column of the first row a leading 1.

1. \( R_1 \leftarrow R_1 / 2 \)
\[
\begin{pmatrix}
1 & -1.5 & -2.5 & | & 0.5 \\
3 & 1 & -2 & | & -4 \\
1 & -2 & 1 & | & 5 \\
\end{pmatrix}
\]

**Step 3: Eliminate the first variable \(x\) from the second and third rows.**

2. \( R_2 \leftarrow R_2 - 3R_1 \) and \( R_3 \leftarrow R_3 - R_1 \)
\[
\begin{pmatrix}
1 & -1.5 & -2.5 & | & 0.5 \\
0 & 5.5 & 5.5 & | & -5.5 \\
0 & -0.5 & 3.5 & | & 4.5 \\
\end{pmatrix}
\]

**Step 4: Make the coefficient of \(y\) in the second row a leading 1.**

3. \( R_2 \leftarrow R_2 / 5.5 \)
\[
\begin{pmatrix}
1 & -1.5 & -2.5 & | & 0.5 \\
0 & 1 & 1 & | & -1 \\
0 & -0.5 & 3.5 & | & 4.5 \\
\end{pmatrix}
\]

**Step 5: Eliminate the second variable \(y\) from the third row.**

4. \( R_3 \leftarrow R_3 + 0.5R_2 \)
\[
\begin{pmatrix}
1 & -1.5 & -2.5 & | & 0.5 \\
0 & 1 & 1 & | & -1 \\
0 & 0 & 4 & | & 4 \\
\end{pmatrix}
\]

**Step 6: Solve the triangular system.**

Now we can solve this system using back substitution.

From \(R_3\):
\[
4z = 4 \implies z = 1
\]

From \(R_2\):
\[
y + z = -1 \implies y + 1 = -1 \implies y = -2
\]

From \(R_1\):
\[
x - 1.5y - 2.5z = 0.5 \implies x - 1.5(-2) - 2.5(1) = 0.5 \implies x + 3 - 2.5 = 0.5 \implies x = 0
\]

**Solution:**
\[
x = 0, \quad y = -2, \quad z = 1
\]

This is the solution to the system of equations.

Do you want any further details or have any questions?

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Here are 5 related questions to consider:
1. How would you solve the system using the Gauss-Jordan elimination method?
2. What changes in the Gaussian elimination process when the system has infinite solutions?
3. Can Gaussian elimination be used for systems with no solutions?
4. How would you solve the system if it were a 4x4 matrix instead?
5. What are the advantages of using Gaussian elimination over substitution?

**Tip:** Gaussian elimination is more efficient for large systems of equations than substitution or elimination methods by hand.

Solve the system of equations: 2x - 3y - 5z = 1, 3x + y - 2z = -4, x - 2y + z = 5 using Gaussian method.

Explore how to solve the system of linear equations 2x - 3y - 5z = 1, 3x + y - 2z = -4, x - 2y + z = 5 using the Gaussian elimination method. Learn key linear algebra concepts, row operations, and back substitution techniques. Suitable for students in Grades 10-12.

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