12-240/Classnotes for Tuesday November 15: Difference between revisions

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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>
|}
|}
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>

Latest revision as of 20:42, 15 November 2012

Problem. Find the rank the matrix

[math]\displaystyle{ 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].

Solution. Using (invertible!) row/column operations we aim to bring [math]\displaystyle{ A }[/math] to look as close as possible to an identity matrix:

Do Get
1. Bring a [math]\displaystyle{ 1 }[/math] to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by [math]\displaystyle{ 1/4 }[/math]. [math]\displaystyle{ \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]\displaystyle{ (-8) }[/math] times the first row to the third row, in order to cancel the [math]\displaystyle{ 8 }[/math] in position 3-1. [math]\displaystyle{ \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]
3. Likewise add [math]\displaystyle{ (-6) }[/math] times the first row to the fourth row, in order to cancel the [math]\displaystyle{ 6 }[/math] in position 4-1. [math]\displaystyle{ \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). [math]\displaystyle{ \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]
5. Turn the 2-2 entry to a [math]\displaystyle{ 1 }[/math] by multiplying the second row by [math]\displaystyle{ 1/2 }[/math]. [math]\displaystyle{ \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]\displaystyle{ 1 }[/math] at position 2-2. [math]\displaystyle{ \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]
7. Using three column operations clean the second row except the pivot. [math]\displaystyle{ \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]\displaystyle{ 4 }[/math] in position 3-3 by first multiplying the third row by [math]\displaystyle{ 1/4 }[/math] and then performing the appropriate row and column transformations. Notice that by pure luck, the [math]\displaystyle{ 4 }[/math] at position 4-5 of the matrix gets killed in action. [math]\displaystyle{ \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.


Lecture notes scanned by Zetalda