Math  /  Algebra

Question|EXERCISE SET 1.2
Reduced Echelon Form of a Matrix
1. Determine whether the following matrices are in reduced echelon form. If a matrix is not in reduced echelon form give a reason.

(a) [103200180149]\left[\begin{array}{rrrr}1 & 0 & 3 & -2 \\ 0 & 0 & 1 & 8 \\ 0 & 1 & 4 & 9\end{array}\right] (b) [120040010600015]\left[\begin{array}{lllll}1 & 2 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & 5\end{array}\right] [10426012340001200001]\left[\begin{array}{lllll}1 & 0 & 4 & 2 & 6 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 1\end{array}\right] (e) [10203000000120700013]\left[\begin{array}{lllll}1 & 0 & 2 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 & 7 \\ 0 & 0 & 0 & 1 & 3\end{array}\right] (f) [10400012000001000001]\left[\begin{array}{lllll}1 & 0 & 4 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\end{array}\right] (g) [100530010301237]\left[\begin{array}{lllll}1 & 0 & 0 & 5 & 3 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 1 & 2 & 3 & 7\end{array}\right] (h) [001040001501003]\left[\begin{array}{lllll}0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \\ 0 & 1 & 0 & 0 & 3\end{array}\right] (i) [153070001400000]\left[\begin{array}{rrrrr}1 & 5 & -3 & 0 & 7 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]

Studdy Solution

STEP 1

1. A matrix is in reduced row echelon form (RREF) if it satisfies the following conditions: - Each leading entry (the first non-zero number from the left, in a row) is 1. - Each leading 1 is the only non-zero entry in its column. - The leading 1 of a given row is to the right of the leading 1 in the row above. - Any row containing only zeros is at the bottom of the matrix.

STEP 2

1. Review each matrix to determine if it satisfies the RREF conditions.
2. Provide a reason if a matrix is not in RREF.

STEP 3

Matrix (a): [103200180149]\left[\begin{array}{rrrr} 1 & 0 & 3 & -2 \\ 0 & 0 & 1 & 8 \\ 0 & 1 & 4 & 9 \end{array}\right]
- The leading 1 in the second row is in the third column, but the leading 1 in the third row is in the second column. This violates the rule that each leading 1 must be to the right of the leading 1 in the row above.

STEP 4

Matrix (b): [120040010600015]\left[\begin{array}{lllll} 1 & 2 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & 5 \end{array}\right]
- The leading 1s are in the correct positions and are the only non-zero entries in their columns. The matrix satisfies all conditions for RREF.

STEP 5

Matrix (c): [10426012340001200001]\left[\begin{array}{lllll} 1 & 0 & 4 & 2 & 6 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right]
- The leading 1s are in the correct positions and are the only non-zero entries in their columns. The matrix satisfies all conditions for RREF.

STEP 6

Matrix (e): [10203000000120700013]\left[\begin{array}{lllll} 1 & 0 & 2 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 & 7 \\ 0 & 0 & 0 & 1 & 3 \end{array}\right]
- The second row is a zero row and is correctly placed at the bottom. However, the leading 1 in the third row is not the only non-zero entry in its column (column 2 has a non-zero entry in the first row).

STEP 7

Matrix (f): [10400012000001000001]\left[\begin{array}{lllll} 1 & 0 & 4 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right]
- The leading 1s are in the correct positions and are the only non-zero entries in their columns. The matrix satisfies all conditions for RREF.

STEP 8

Matrix (g): [100530010301237]\left[\begin{array}{lllll} 1 & 0 & 0 & 5 & 3 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 1 & 2 & 3 & 7 \end{array}\right]
- The leading 1 in the third row is in the second column, which is not to the right of the leading 1 in the second row. This violates the RREF condition.

STEP 9

Matrix (h): [001040001501003]\left[\begin{array}{lllll} 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \\ 0 & 1 & 0 & 0 & 3 \end{array}\right]
- The leading 1 in the third row is in the second column, which is not to the right of the leading 1 in the second row. This violates the RREF condition.

STEP 10

Matrix (i): [153070001400000]\left[\begin{array}{rrrrr} 1 & 5 & -3 & 0 & 7 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]
- The leading 1s are in the correct positions and are the only non-zero entries in their columns. The matrix satisfies all conditions for RREF.

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