Difference between revisions of "12240/Classnotes for Tuesday November 15"
From Drorbn
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{ border="1px" cellspadding="5" cellspacing=0 style="fontsize:90%;"  { border="1px" cellspadding="5" cellspacing=0 style="fontsize:90%;"  
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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 31.  2. Add <math>(8)</math> times the first row to the third row, in order to cancel the <math>8</math> in position 31.  
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 41.  3. Likewise add <math>(6)</math> times the first row to the fourth row, in order to cancel the <math>6</math> in position 41.  
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 22 entry to a <math>1</math> by multiplying the second row by <math>1/2</math>.  5. Turn the 22 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 22.  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 22.  
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 33 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 45 of the matrix gets killed in action.  8. Clean up the row and the column of the <math>4</math> in position 33 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 45 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:ZetaldaZetalda]] ==  
+  <gallery>  
+  Image:12240Nov151.jpegPage 1  
+  Image:12240Nov152.jpegPage 2  
+  </gallery> 
Latest revision as of 21:42, 15 November 2012

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 31.  
3. Likewise add times the first row to the fourth row, in order to cancel the in position 41.  
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 22 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 22.  
7. Using three column operations clean the second row except the pivot.  
8. Clean up the row and the column of the in position 33 by first multiplying the third row by and then performing the appropriate row and column transformations. Notice that by pure luck, the at position 45 of the matrix gets killed in action. 
Thus the rank of our matrix is 3.