Math Problem Statement

Which of the following matrices is/are in reduced row-echelon form

$A=\begin{pmatrix}1&2&0\0&1&0 \0&0&1\end{pmatrix}$,

$B=\begin{pmatrix}0&1&0\1&0&0\0&0&1\end{pmatrix}$,

$C=\begin{pmatrix}1&0&0\0&2&0\0&0&1\end{pmatrix}$,

$D=\begin{pmatrix}1&2&0\0&0&1\0&0&0\0&0&0\end{pmatrix}$?

Solution

To determine which matrices are in reduced row-echelon form (RREF), we need to verify that each matrix satisfies the following criteria:

  1. Leading entry: The leading entry (first non-zero number from the left) in each non-zero row is 1.
  2. Column zeroes: Each leading 1 is the only non-zero entry in its column.
  3. Row order: The leading 1 in a row is to the right of the leading 1 in the row above it.
  4. Row of zeroes: Any row consisting entirely of zeroes is at the bottom of the matrix.

Let's analyze each matrix:

Matrix AA:

A=(120010001)A = \begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}

  1. Leading entries are 1.
  2. Leading 1's are the only non-zero entries in their columns.
  3. Each leading 1 is to the right of the leading 1 in the row above it.
  4. No row of all zeroes.

Matrix AA is in RREF.

Matrix BB:

B=(010100001)B = \begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}

  1. The first row has a leading 1, but it's not in the first column.
  2. Leading 1's are the only non-zero entries in their columns.
  3. The leading 1 in the second row is not to the right of the leading 1 in the first row.

Matrix BB is not in RREF.

Matrix CC:

C=(100020001)C = \begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix}

  1. The first row has a leading 1.
  2. The second row has a leading entry of 2 (not 1), which violates the first condition.
  3. The third row is fine but the violation in the second row is enough.

Matrix CC is not in RREF.

Matrix DD:

D=(120001000000)D = \begin{pmatrix}1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}

  1. Leading entries are 1.
  2. Leading 1's are the only non-zero entries in their columns.
  3. Each leading 1 is to the right of the leading 1 in the row above it.
  4. Rows of all zeroes are at the bottom.

Matrix DD is in RREF.

Conclusion

The matrices in reduced row-echelon form are:

A=(120010001)A = \begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} D=(120001000000)D = \begin{pmatrix}1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}

Would you like further details or have any questions?

Here are 5 questions you might ask next:

  1. How do you perform Gaussian elimination to achieve RREF?
  2. What is the difference between row-echelon form and reduced row-echelon form?
  3. Can you provide an example of a matrix not in RREF and show how to convert it?
  4. What are the applications of reduced row-echelon form in solving linear systems?
  5. How do you determine the rank of a matrix from its RREF?

Tip: In reduced row-echelon form, each leading 1 must be the only non-zero entry in its column, ensuring that each row and column maintains a specific order.

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Math Problem Analysis

Mathematical Concepts

Matrices
Row Operations
Reduced Row-Echelon Form (RREF)

Formulas

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Theorems

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Suitable Grade Level

Advanced High School