Consider the matrix: 1 2 -2 0 -1 3 4 1-1, Perform R3 - 2R2: 1 2 0 0 3 2 4 3

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**Matrix Row Operation Example**

Consider the matrix:

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

To perform the row operation \( R3 - 2R2 \), follow these steps:

1. Identify and modify the third row (\(R3\)) by subtracting two times the second row (\(2R2\)) from it.

2. The matrix becomes:

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

**Explanation:**

- The first row remains unchanged: \( (1, 0, -1) \).

- The operation \( R3 - 2R2 \) is applied to the third row:
  - The first element: \(-2 - 2(2) = 0\).
  - The second element: \(1 - 2(3) = 2\).
  - The third element: \(-1 - 2(4) = 3\).

- The second row \( (2, 3, 4) \) remains unchanged.

This procedure illustrates how basic row operations can transform matrices in linear algebra.
Transcribed Image Text:**Matrix Row Operation Example** Consider the matrix: \[ \begin{pmatrix} 1 & 0 & -1 \\ 2 & 3 & 4 \\ -2 & 1 & -1 \\ \end{pmatrix} \] To perform the row operation \( R3 - 2R2 \), follow these steps: 1. Identify and modify the third row (\(R3\)) by subtracting two times the second row (\(2R2\)) from it. 2. The matrix becomes: \[ \begin{pmatrix} 1 & 0 & -1 \\ 2 & 3 & 4 \\ 0 & 2 & 3 \\ \end{pmatrix} \] **Explanation:** - The first row remains unchanged: \( (1, 0, -1) \). - The operation \( R3 - 2R2 \) is applied to the third row: - The first element: \(-2 - 2(2) = 0\). - The second element: \(1 - 2(3) = 2\). - The third element: \(-1 - 2(4) = 3\). - The second row \( (2, 3, 4) \) remains unchanged. This procedure illustrates how basic row operations can transform matrices in linear algebra.
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