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how to solve this one using the gaussian elimination?

To solve this system of linear equations using **Gaussian elimination**, the goal is to convert the system into an upper triangular form, then back-substitute to find the values of \( x \), \( y \), and \( z \).

The system of equations is:

\[
\begin{aligned}
1) & \quad 2x - 3y - 9z = -5 \\
2) & \quad x + 3z = 2 \\
3) & \quad -3x + y - 4z = -3
\end{aligned}
\]

### Step 1: Write the augmented matrix for the system.

\[
\begin{pmatrix}
2 & -3 & -9 & | & -5 \\
1 & 0  & 3  & | & 2 \\
-3 & 1 & -4 & | & -3
\end{pmatrix}
\]

### Step 2: Perform row operations to get an upper triangular matrix.

We aim to create zeros below the diagonal.

#### a) Make the first element of row 2 zero:
We already have a 1 in the second row, first column (i.e., in row 2, column 1), so we can eliminate the first element in row 1. Use **Row 1 = Row 1 - 2 × Row 2**.

\[
\text{Row 1: } \begin{pmatrix} 2 & -3 & -9 & | & -5 \end{pmatrix} - 2 \times \begin{pmatrix} 1 & 0 & 3 & | & 2 \end{pmatrix}
\]

New row 1:
\[
\begin{pmatrix} 0 & -3 & -15 & | & -9 \end{pmatrix}
\]

The matrix now becomes:
\[
\begin{pmatrix}
0 & -3 & -15 & | & -9 \\
1 & 0 & 3 & | & 2 \\
-3 & 1 & -4 & | & -3
\end{pmatrix}
\]

#### b) Eliminate the first element in row 3:
Use **Row 3 = Row 3 + 3 × Row 2** to eliminate the \(-3\) in the first column of row 3.

\[
\text{Row 3: } \begin{pmatrix} -3 & 1 & -4 & | & -3 \end{pmatrix} + 3 \times \begin{pmatrix} 1 & 0 & 3 & | & 2 \end{pmatrix}
\]

New row 3:
\[
\begin{pmatrix} 0 & 1 & 5 & | & 3 \end{pmatrix}
\]

The updated matrix becomes:
\[
\begin{pmatrix}
0 & -3 & -15 & | & -9 \\
1 & 0 & 3 & | & 2 \\
0 & 1 & 5 & | & 3
\end{pmatrix}
\]

### Step 3: Scale the second row:
To make the leading coefficient in the second row equal to 1, divide the entire second row by \(-3\).

\[
\text{Row 1: } \begin{pmatrix} 0 & -3 & -15 & | & -9 \end{pmatrix} \Rightarrow \begin{pmatrix} 0 & 1 & 5 & | & 3 \end{pmatrix}
\]

Updated matrix:
 

How to solve this one using Gaussian elimination?

Learn how to solve a system of three linear equations using Gaussian elimination. The system involves variables x, y, and z, and the solution uses matrix transformations and row operations to simplify the system into upper triangular form for back-substitution.

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