Complete the reduced row echelon form for the matrix shown.

⎡⎣⎢3402−11101∣∣∣∣16−511⎤⎦⎥[3402−11101|16−511]

⎡⎣⎢100a100c1∣∣∣∣bde⎤⎦

To solve for the complete reduced row echelon form (RREF) of the matrix 

\[
\left[\begin{array}{ccc|ccc}
3 & 4 & 0 & 2 & -1 & 1 \\
1 & 1 & 0 & 1 & 6 & -5 \\
1 & 1 & 0 & 1 & 1 & 1
\end{array}\right]
\]

and transform it into the following form:

\[
\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & a & b & c \\
0 & 1 & 0 & d & e & f \\
0 & 0 & 1 & g & h & i
\end{array}\right]
\]

we will perform row operations to transform the matrix into reduced row echelon form. Let's go through the process step by step.

### Step 1: Create a leading 1 in the top-left corner (row 1, column 1)
We can start by dividing row 2 by 1 (since row 2 already has a leading 1 in column 1).

\[
R_2 = R_2
\]

Thus, the matrix becomes:

\[
\left[\begin{array}{ccc|ccc}
3 & 4 & 0 & 2 & -1 & 1 \\
1 & 1 & 0 & 1 & 6 & -5 \\
1 & 1 & 0 & 1 & 1 & 1
\end{array}\right]
\]

### Step 2: Eliminate non-zero entries in column 1 for rows 1 and 3
We can subtract appropriate multiples of row 2 from rows 1 and 3 to eliminate the entries in column 1.

- \( R_1 \leftarrow R_1 - 3R_2 \)
- \( R_3 \leftarrow R_3 - R_2 \)

After these operations, the matrix becomes:

\[
\left[\begin{array}{ccc|ccc}
0 & 1 & 0 & -1 & 19 & -14 \\
1 & 1 & 0 & 1 & 6 & -5 \\
0 & 0 & 0 & 0 & -5 & 6
\end{array}\right]
\]

### Step 3: Create a leading 1 in the second column (row 2, column 2)
We already have a leading 1 in row 2, column 2. Now, we just need to subtract row 2 from row 1 to eliminate the entry in row 1, column 2.

- \( R_1 \leftarrow R_1 - R_2 \)

This gives us the following matrix:

\[
\left[\begin{array}{ccc|ccc}
0 & 0 & 0 & -2 & 13 & -9 \\
1 & 1 & 0 & 1 & 6 & -5 \\
0 & 0 & 0 & 0 & -5 & 6
\end{array}\right]
\]

### Step 4: Handle the third row
It appears that we have a zero row in the third row, so we need to move forward with other operations.

Thus, the reduced row echelon form of the matrix is:

\[
\left[\begin{array}{ccc|ccc}
0 & 0 & 0 & -2 & 13 & -9 \\
1 & 1 & 0 & 1 & 6 & -5 \\
0 & 0 & 0 & 0 & -5 & 6
\end{array}\right]
\]

Let me know if you'd like more detailed explanations or have questions.

### 5 Related Questions:
1. What are the conditions for a matrix to be in reduced row echelon form?
2. How do row operations affect the solution of a system of equations?
3. What does it mean when a row of all zeros appears in a matrix during row reduction?
4. Can every matrix be transformed into reduced row echelon form? Why?
5. How does reduced row echelon form relate to solving linear systems?

### Tip:
When reducing matrices, keep track of row operations carefully to avoid mistakes—writing them down step-by-step can help ensure accuracy.

Complete the reduced row echelon form for the matrix shown: ⎡⎣⎢3402−11101∣∣∣∣16−511⎤⎦⎥.

Learn how to perform row operations to obtain the reduced row echelon form (RREF) for the matrix [3402−11101|16−511]. This problem demonstrates matrix transformations using Gaussian elimination and is suitable for students in Grades 10-12 or college-level linear algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation