Talk:06-240/Classnotes For Tuesday November 14: Difference between revisions
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Reduced row echelon form - Is there a reason to make column with entry 1 to the form of e<sub>n</sub> (1 at n<sup>th</sup> row, 0 for all other entries)? According to some books, matrix <math>\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}</math> is good enough to show that the rank of the matrix is 3. This is because the first three rows are linearly independent, they can't form linear combination for preceeding rows. Anyone could |
Reduced row echelon form - Is there a reason to make column with entry 1 to the form of e<sub>n</sub> (1 at n<sup>th</sup> row, 0 for all other entries)? According to some books, matrix <math>\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}</math> is good enough to show that the rank of the matrix is 3. This is because the first three rows are linearly independent, they can't form linear combination for preceeding rows. Anyone could please explain why we have to reduce it to <math>\begin{pmatrix}1&0&-4&0&-2\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}</math>? Thank you. [[User:Wongpak|Wongpak]] 23:54, 15 November 2006 (EST) |
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Revision as of 23:55, 15 November 2006
Reduced row echelon form - Is there a reason to make column with entry 1 to the form of en (1 at nth row, 0 for all other entries)? According to some books, matrix is good enough to show that the rank of the matrix is 3. This is because the first three rows are linearly independent, they can't form linear combination for preceeding rows. Anyone could please explain why we have to reduce it to ? Thank you. Wongpak 23:54, 15 November 2006 (EST)
