x-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 the system:
\[
\begin{aligned}
1. \quad & x - 3y - 5z = 1 \\
2. \quad & 3x + y - 2z = -4 \\
3. \quad & x - 2y + z = 5
\end{aligned}
\]

We represent the system as an augmented matrix:

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

### Step 1: Make the pivot in the first column equal to 1 (already done here for the first row).

Now, let's **eliminate the x-terms** from rows 2 and 3 using row operations:

- Row 2: \( R_2 \rightarrow R_2 - 3R_1 \)
- Row 3: \( R_3 \rightarrow R_3 - R_1 \)

Perform these operations:

\[
\begin{aligned}
R_2 & \rightarrow (3 \quad 1 \quad -2 \quad | \quad -4) - 3(1 \quad -3 \quad -5 \quad | \quad 1) \\
& = (3 - 3 \quad 1 - (-9) \quad -2 - (-15) \quad | \quad -4 - 3) \\
& = (0 \quad 10 \quad 13 \quad | \quad -7) \\
\\
R_3 & \rightarrow (1 \quad -2 \quad 1 \quad | \quad 5) - (1 \quad -3 \quad -5 \quad | \quad 1) \\
& = (1 - 1 \quad -2 - (-3) \quad 1 - (-5) \quad | \quad 5 - 1) \\
& = (0 \quad 1 \quad 6 \quad | \quad 4)
\end{aligned}
\]

The new matrix is:
\[
\begin{pmatrix}
1 & -3 & -5 & | & 1 \\
0 & 10 & 13 & | & -7 \\
0 & 1 & 6 & | & 4
\end{pmatrix}
\]

### Step 2: Make the pivot in the second column equal to 1 by dividing the second row by 10.

- Row 2: \( R_2 \rightarrow \frac{R_2}{10} \)

\[
\begin{pmatrix}
1 & -3 & -5 & | & 1 \\
0 & 1 & 1.3 & | & -0.7 \\
0 & 1 & 6 & | & 4
\end{pmatrix}
\]

### Step 3: Eliminate the second column from the third row using row operations:

- Row 3: \( R_3 \rightarrow R_3 - R_2 \)

Perform the operation:
\[
R_3 \rightarrow (0 \quad 1 \quad 6 \quad | \quad 4) - (0 \quad 1 \quad 1.3 \quad | \quad -0.7)
= (0 \quad 0 \quad 4.7 \quad | \quad 4.7)
\]

The new matrix is:
\[
\begin{pmatrix}
1 & -3 & -5 & | & 1 \\
0 & 1 & 1.3 & | & -0.7 \\
0 & 0 & 4.7 & | & 4.7
\end{pmatrix}
\]

### Step 4: Make the pivot in the third column equal to 1 by dividing the third row by 4.7:

\[
R_3 \rightarrow \frac{R_3}{4.7} = \left( 0 \quad 0 \quad 1 \quad | \quad 1 \right)
\]

The matrix becomes:
\[
\begin{pmatrix}
1 & -3 & -5 & | & 1 \\
0 & 1 & 1.3 & | & -0.7 \\
0 & 0 & 1 & | & 1
\end{pmatrix}
\]

### Step 5: Back-substitution to solve for variables:

- From \( R_3 \): \( z = 1 \)
- From \( R_2 \): \( y + 1.3z = -0.7 \implies y + 1.3(1) = -0.7 \implies y = -2 \)
- From \( R_1 \): \( x - 3y - 5z = 1 \implies x - 3(-2) - 5(1) = 1 \implies x + 6 - 5 = 1 \implies x = 0 \)

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

Would you like more details or have any questions about this method?

Here are 5 related questions to explore:

1. How does the Gaussian elimination differ from Gauss-Jordan elimination?
2. What are the necessary conditions for a system of equations to have a unique solution?
3. How can Gaussian elimination be applied to larger systems, such as 4x4 matrices?
4. What are the implications if the matrix contains a row of zeros during the elimination process?
5. How can Gaussian elimination be used to find the inverse of a matrix?

**Tip**: Gaussian elimination can also be used to determine if a system of equations has no solution or infinitely many solutions by analyzing the row-reduced form of the matrix.

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

Step-by-step guide to solving the system of equations x - 3y - 5z = 1, 3x + y - 2z = -4, x - 2y + z = 5 using the Gaussian elimination method. Learn how to reduce the system to row echelon form and use back-substitution to find the solution: x = 0, y = -2, z = 1. Suitable for students in Grades 10-12.

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