12-240/Classnotes for Tuesday November 15

DBN: Classes: 2012-13: 12-240 / Navigation Additions to this web site no longer count towards good deed points. # Week of... Notes and Links 1 Sep 10 About This Class, Tuesday, Thursday 2 Sep 17 HW1, Tuesday, Thursday, HW1 Solutions 3 Sep 24 HW2, Tuesday, Class Photo, Thursday 4 Oct 1 HW3, Tuesday, Thursday 5 Oct 8 HW4, Tuesday, Thursday 6 Oct 15 Tuesday, Thursday 7 Oct 22 HW5, Tuesday, Term Test was on Thursday. HW5 Solutions 8 Oct 29 Why LinAlg?, HW6, Tuesday, Thursday, Nov 4 is the last day to drop this class 9 Nov 5 Tuesday, Thursday 10 Nov 12 Monday-Tuesday is UofT November break, HW7, Thursday 11 Nov 19 HW8, Tuesday,Thursday 12 Nov 26 HW9, Tuesday , Thursday 13 Dec 3 Tuesday UofT Fall Semester ends Wednesday F Dec 10 The Final Exam (time, place, style, office hours times) Register of Good Deeds Add your name / see who's in!

Problem. Find the rank the matrix

$A={\begin{pmatrix}0&2&4&2&2\\4&4&4&8&0\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}}$.

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

Do	Get
1. Bring a $1$ to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by $1/4$.	${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}}$
2. Add $(-8)$ times the first row to the third row, in order to cancel the $8$ in position 3-1.	${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}}$
3. Likewise add $(-6)$ times the first row to the fourth row, in order to cancel the $6$ in position 4-1.	${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}$
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).	${\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}$
5. Turn the 2-2 entry to a $1$ by multiplying the second row by $1/2$.	${\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}$
6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" $1$ at position 2-2.	${\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}$
7. Using three column operations clean the second row except the pivot.	${\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}$
8. Clean up the row and the column of the $4$ in position 3-3 by first multiplying the third row by $1/4$ and then performing the appropriate row and column transformations. Notice that by pure luck, the $4$ at position 4-5 of the matrix gets killed in action.	${\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}}$

Thus the rank of our matrix is 3.

Lecture notes scanned by Zetalda

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