12-240/Classnotes for Tuesday November 15: Difference between revisions
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|align=center|'''Do''' |
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|align=center|'''Get''' |
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|align=center|'''Do''' |
|align=center|'''Do''' |
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|align=center|'''Get''' |
|align=center|'''Get''' |
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|1. Bring a <math>1</math> to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by <math>1/4</math>. |
|1. Bring a <math>1</math> to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by <math>1/4</math>. |
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|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}</math> |
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|2. Add <math>(-8)</math> times the first row to the third row, in order to cancel the <math>8</math> in position 3-1. |
|2. Add <math>(-8)</math> times the first row to the third row, in order to cancel the <math>8</math> in position 3-1. |
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|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}</math> |
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|3. Likewise add <math>(-6)</math> times the first row to the fourth row, in order to cancel the <math>6</math> in position 4-1. |
|3. Likewise add <math>(-6)</math> times the first row to the fourth row, in order to cancel the <math>6</math> in position 4-1. |
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|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math> |
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|4. With similar column operations (you need three of those) cancel all the entries in the first row (except, of course, the first, which is used in the canceling). |
|4. With similar column operations (you need three of those) cancel all the entries in the first row (except, of course, the first, which is used in the canceling). |
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|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math> |
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|5. Turn the 2-2 entry to a <math>1</math> by multiplying the second row by <math>1/2</math>. |
|5. Turn the 2-2 entry to a <math>1</math> by multiplying the second row by <math>1/2</math>. |
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|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math> |
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|6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" <math>1</math> at position 2-2. |
|6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" <math>1</math> at position 2-2. |
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|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}</math> |
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|7. Using three column operations clean the second row except the pivot. |
|7. Using three column operations clean the second row except the pivot. |
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|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}</math> |
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|8. Clean up the row and the column of the <math>4</math> in position 3-3 by first multiplying the third row by <math>1/4</math> and then performing the appropriate row and column transformations. Notice that by pure luck, the <math>4</math> at position 4-5 of the matrix gets killed in action. |
|8. Clean up the row and the column of the <math>4</math> in position 3-3 by first multiplying the third row by <math>1/4</math> and then performing the appropriate row and column transformations. Notice that by pure luck, the <math>4</math> at position 4-5 of the matrix gets killed in action. |
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|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}</math> |
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}</math> |
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Thus the rank of our matrix is 3. |
Thus the rank of our matrix is 3. |
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== Lecture notes scanned by [[User:Zetalda|Zetalda]] == |
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<gallery> |
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Image:12-240-Nov15-1.jpeg|Page 1 |
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Image:12-240-Nov15-2.jpeg|Page 2 |
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</gallery> |
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Latest revision as of 21:42, 15 November 2012
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Problem. Find the rank the matrix
Solution. Using (invertible!) row/column operations we aim to bring to look as close as possible to an identity matrix:
| Do | Get |
| 1. Bring a to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by . | |
| 2. Add times the first row to the third row, in order to cancel the in position 3-1. | |
| 3. Likewise add times the first row to the fourth row, in order to cancel the in position 4-1. | |
| 4. With similar column operations (you need three of those) cancel all the entries in the first row (except, of course, the first, which is used in the canceling). | |
| 5. Turn the 2-2 entry to a by multiplying the second row by . | |
| 6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" at position 2-2. | |
| 7. Using three column operations clean the second row except the pivot. | |
| 8. Clean up the row and the column of the in position 3-3 by first multiplying the third row by and then performing the appropriate row and column transformations. Notice that by pure luck, the at position 4-5 of the matrix gets killed in action. |
Thus the rank of our matrix is 3.
Lecture notes scanned by Zetalda
Page 1
Page 2
