Math Problem Statement
Find the rank of the matrix according to row echelon form conditions
Solution
To find the rank of the matrix, we need to perform row operations to bring the matrix into row echelon form (REF). The rank of the matrix is the number of non-zero rows in its REF.
Let’s start by analyzing the matrix given in the image:
1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ \end{pmatrix}$$ ### Step 1: Convert to Row Echelon Form 1. **First Pivot (Column 1)**: - The first row already has a leading 1 in the first position. Subtract the first row from all the rows below it to make the first column below the pivot all zeros: $$R_2 \rightarrow R_2 - R_1, \quad R_3 \rightarrow R_3 - R_1, \quad R_4 \rightarrow R_4 - R_1, \quad R_5 \rightarrow R_5 - R_1$$ This results in: $$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & -1 & -1 & -1 \\ 0 & -1 & 0 & -1 & -1 \\ 0 & -1 & -1 & 0 & -1 \\ 0 & -1 & -1 & -1 & 0 \\ \end{pmatrix}$$ 2. **Second Pivot (Column 2)**: - The second row does not have a leading 1, but the third row has a leading -1 in the second position, so we can swap row 2 and row 3 to get a leading 1 after normalization: $$R_2 \leftrightarrow R_3, \quad \text{then normalize } R_2 \text{ by multiplying by } -1$$ Result: $$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & -1 & -1 & -1 \\ 0 & -1 & -1 & 0 & -1 \\ 0 & -1 & -1 & -1 & 0 \\ \end{pmatrix}$$ Next, subtract appropriate multiples of the second row from the rows below it to clear out the second column below the pivot: $$R_3 \rightarrow R_3 + R_2, \quad R_4 \rightarrow R_4 + R_2, \quad R_5 \rightarrow R_5 + R_2$$ Result: $$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 & 1 \\ \end{pmatrix}$$ 3. **Third Pivot (Column 3)**: - Normalize the third row by multiplying by -1 to get a leading 1: $$R_3 \rightarrow -R_3$$ Result: $$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 & 1 \\ \end{pmatrix}$$ Then, add the third row to the rows below it to clear out the third column below the pivot: $$R_4 \rightarrow R_4 + R_3, \quad R_5 \rightarrow R_5 + R_3$$ Result: $$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix}$$ ### Step 2: Determine the Rank In the final row echelon form, each row has a leading 1 in a distinct column, and there are 5 such non-zero rows. Thus, **the rank of the matrix is 5**. Would you like more details or have any questions? ### Further Questions: 1. What is the general method to determine the rank of a matrix? 2. Can you explain the significance of the rank of a matrix in linear algebra? 3. How does the row echelon form differ from the reduced row echelon form? 4. What is the maximum rank a matrix can have given its dimensions? 5. How would the rank change if we added a row of zeros to the matrix? ### Tip: Always ensure that the matrix is fully simplified to row echelon form before counting the number of non-zero rows to determine the rank.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Operations
Row Echelon Form
Linear Algebra
Formulas
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Theorems
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Suitable Grade Level
Advanced High School or College