12-240/Classnotes for Tuesday November 15 - Revision history

Zetalda at 01:42, 16 November 2012

2012-11-16T01:42:32Z

← Older revision		Revision as of 21:42, 15 November 2012
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	\|}		\|}
	Thus the rank of our matrix is 3.		Thus the rank of our matrix is 3.


			== Lecture notes scanned by [[User:Zetalda\|Zetalda]] ==
			<gallery>
			Image:12-240-Nov15-1.jpeg\|Page 1
			Image:12-240-Nov15-2.jpeg\|Page 2
			</gallery>

Drorbn at 19:27, 15 November 2012

2012-11-15T19:27:20Z

← Older revision		Revision as of 15:27, 15 November 2012
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	{\| border="1px" cellspadding="5" cellspacing=0 style="font-size:90%;"		{\| border="1px" cellspadding="5" cellspacing=0 style="font-size:90%;"
	\|+		\|+
	\|align=center\|'''Do'''
	\|align=center\|'''Get'''
	\|align=center\|'''Do'''		\|align=center\|'''Do'''
	\|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>.
	\|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>
			\|- valign=top
	\|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.
	\|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.
	\|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>
			\|- valign=top
	\|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).
	\|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>.
	\|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>
			\|- valign=top
	\|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.
	\|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.
	\|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>
			\|- valign=top
	\|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.
	\|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>

Drorbn: 12-240/Classnotes for Tuesday November 8 moved to 12-240/Classnotes for Tuesday November 15

2012-11-14T19:27:36Z

12-240/Classnotes for Tuesday November 8 moved to 12-240/Classnotes for Tuesday November 15

← Older revision	Revision as of 15:27, 14 November 2012
(No difference)

Drorbn at 18:00, 8 November 2012

2012-11-08T18:00:06Z

New page

{{12-240/Navigation}}
'''Problem.''' Find the rank the matrix
<center><math>A=\begin{pmatrix}0&2&4&2&2\\4&4&4&8&0\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}</math>.</center>

'''Solution.''' Using (invertible!) row/column operations we aim to bring <math>A</math> to look as close as possible to an identity matrix:

{| border="1px" cellspadding="5" cellspacing=0 style="font-size:90%;"
|+
|align=center|'''Do'''
|align=center|'''Get'''
|align=center|'''Do'''
|align=center|'''Get'''
|- valign=top
|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>.
|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>
|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.
|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>
|- valign=top
|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.
|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>
|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).
|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>
|- valign=top
|5. Turn the 2-2 entry to a <math>1</math> by multiplying the second row by <math>1/2</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>
|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.
|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>
|- valign=top
|7. Using three column operations clean the second row except the pivot.
|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>
|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.
|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>
|}
Thus the rank of our matrix is 3.