- #1
bwpbruce
- 60
- 1
Please check my solution.View image: Possible Echelon Form of Matrix
How did you make the conclusion that only the first possibility is possible?bwpbruce said:I realize $\textbf{a}_1$ was not a linearly independent column. So only the 1st possibility is possible.
The claim that allows deducing $\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$ is linearly independent under the assumptions of post #1 is the claim in https://driven2services.com/staging/mh/index.php?threads/13869/. Note that it is incorrect to say "$\mathbf{w}$ is linearly independent". Only a set of vectors (even if the set contains one vector) can be linearly dependent.bwpbruce said:BTW, can you clarify what they are referring to when they say $\textbf{w}$ not in span{$\textbf{u,v}$} is linearly independent?
The possible echelon form of a matrix is a simplified version of the original matrix that has been reduced using elementary row operations. It consists of a series of rows where each successive row has more leading zeros than the previous row.
A matrix is in echelon form if it satisfies two conditions: 1) all non-zero rows are above any rows of all zeros, and 2) the leading coefficient (first non-zero entry) of each row is to the right of the leading coefficient of the row above it.
The three elementary row operations are 1) multiplying a row by a non-zero constant, 2) adding a multiple of one row to another row, and 3) swapping two rows.
Yes, any matrix can be reduced to echelon form using the elementary row operations. However, the resulting echelon form may not be unique as there can be multiple ways to reduce a matrix to echelon form.
Having a matrix in echelon form makes it easier to perform operations such as calculating the determinant, finding the inverse, and solving systems of linear equations. It also provides a simplified representation of the original matrix.