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solve the matrix using gaus jordan method applying elementary row operations

To solve the given matrix using the Gauss-Jordan method, we'll perform elementary row operations to convert the matrix into reduced row echelon form (RREF). The matrix in the image is:

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

### Step-by-Step Gauss-Jordan Elimination:

**Matrix:**
\[
\begin{pmatrix}
-2 & 1 & 3 \\
0 & -1 & 1 \\
1 & -4 & 0
\end{pmatrix}
\]

#### Step 1: Make the leading coefficient (pivot) of the first row 1.
To do this, multiply Row 1 by \(-\frac{1}{2}\):

\[
R_1 \rightarrow \frac{1}{-2} R_1
\]

\[
\begin{pmatrix}
1 & -\frac{1}{2} & -\frac{3}{2} \\
0 & -1 & 1 \\
1 & -4 & 0
\end{pmatrix}
\]

#### Step 2: Make the leading entry of Row 3 a 0.
To eliminate the 1 in the first position of Row 3, subtract Row 1 from Row 3:

\[
R_3 \rightarrow R_3 - R_1
\]

\[
\begin{pmatrix}
1 & -\frac{1}{2} & -\frac{3}{2} \\
0 & -1 & 1 \\
0 & -\frac{7}{2} & \frac{3}{2}
\end{pmatrix}
\]

#### Step 3: Make the leading coefficient (pivot) of Row 2 a 1.
Multiply Row 2 by \(-1\):

\[
R_2 \rightarrow -1 R_2
\]

\[
\begin{pmatrix}
1 & -\frac{1}{2} & -\frac{3}{2} \\
0 & 1 & -1 \\
0 & -\frac{7}{2} & \frac{3}{2}
\end{pmatrix}
\]

#### Step 4: Eliminate the second column entry in Row 1.
To eliminate \(-\frac{1}{2}\) in the second column of Row 1, add \(\frac{1}{2} R_2\) to Row 1:

\[
R_1 \rightarrow R_1 + \frac{1}{2} R_2
\]

\[
\begin{pmatrix}
1 & 0 & -2 \\
0 & 1 & -1 \\
0 & -\frac{7}{2} & \frac{3}{2}
\end{pmatrix}
\]

#### Step 5: Make the leading coefficient (pivot) of Row 3 a 1.
Multiply Row 3 by \(-\frac{2}{7}\):

\[
R_3 \rightarrow -\frac{2}{7} R_3
\]

\[
\begin{pmatrix}
1 & 0 & -2 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{pmatrix}
\]

#### Step 6: Eliminate the third column entries in Rows 1 and 2.
- For Row 1, add \(2 R_3\) to Row 1 to make the third column entry 0.
- For Row 2, add \(R_3\) to Row 2 to make the third column entry 0.

\[
R_1 \rightarrow R_1 + 2 R_3
\]
\[
R_2 \rightarrow R_2 + R_3
\]

\[
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\]

### Final Solution:

The matrix is now in reduced row echelon form (RREF):

\[
\boxed{
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
}
\]

This is the identity matrix, which means the system is solved.

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Would you like further details or have any questions?

Here are some related questions to explore:
1. How does the Gauss-Jordan method differ from Gaussian elimination?
2. What are the types of elementary row operations in matrix solving?
3. Can the Gauss-Jordan method be used for non-square matrices?
4. How can matrices be used to solve systems of linear equations?
5. What is the advantage of using the reduced row echelon form?

**Tip:** Practice identifying pivot elements in each row to streamline solving matrices using the Gauss-Jordan method.

Solve the matrix using Gauss-Jordan method applying elementary row operations.

Learn how to solve a 3x3 matrix using the Gauss-Jordan elimination method. This problem covers key concepts in linear algebra, including row operations and reduced row echelon form. Suitable for advanced high school or college-level students.

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