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Week of...

Notes and Links

1

Sep 11

About, Tue, HW1, Putnam, Thu

2

Sep 18

Tue, HW2, Thu

3

Sep 25

Tue, HW3, Photo, Thu

4

Oct 2

Tue, HW4, Thu

5

Oct 9

Tue, HW5, Thu

6

Oct 16

Why?, Iso, Tue, Thu

7

Oct 23

Term Test, Thu (double)

8

Oct 30

Tue, HW6, Thu

9

Nov 6

Tue, HW7, Thu

10

Nov 13

Tue, HW8, Thu

11

Nov 20

Tue, HW9, Thu

12

Nov 27

Tue, HW10, Thu

13

Dec 4

On the final, Tue, Thu

F

Dec 11

Final: Dec 13 25PM at BN3, Exam Forum

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More about the Wongpak Matrices
In Talk:06240/Classnotes_For_Tuesday_November_14, User:Wongpak asked something about row echelon form and reduced row echelon form, and gave the following matrices as specific examples:
$A_{1}={\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}$


$A_{2}={\begin{pmatrix}1&0&4&0&6\\0&1&2&0&2\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}$

So let us assume row reduction leads us to the systems $A_{1}x=b$ or $A_{2}x=b$. What does it tell us about the solutions? Let us start from the second system:
${\begin{pmatrix}1&0&4&0&6\\0&1&2&0&2\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{pmatrix}}={\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}$

or

$x_{1}$


$4x_{3}$


$6x_{5}$

$=$

$b_{1}$


$x_{2}$

$+2x_{3}$


$2x_{5}$

$=$

$b_{2}$




$x_{4}$

$+2x_{5}$

$=$

$b_{3}$





$0$

$=$

$b_{4}$


Well, quite clearly if $b_{4}\neq 0$ this system has no solutions, but if $b_{4}=0$ it has solutions no matter what $b_{1}$, $b_{2}$ and $b_{3}$ are. Finally, for any given values of $b_{1}$, $b_{2}$ and $b_{3}$ we can choose the values of $x_{3}$ and $x_{5}$ (the variables corresponding the columns containing no pivots) as we please, and then get solutions by setting the "pivotal variables" in terms of the nonpivotal ones as follows: $x_{1}=b_{1}+4x_{3}+6x_{5}$, $x_{2}=b_{2}2x_{3}+2x_{5}$ and $x_{4}=b_{3}2x_{5}$.
What about the system corresponding to $A_{1}$? It is
${\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{pmatrix}}={\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}$

or

$x_{1}$

$+3x_{2}$

$+2x_{3}$

$+4x_{4}$

$+2x_{5}$

$=$

$b_{1}$


$x_{2}$

$+2x_{3}$

$+3x_{4}$

$+4x_{5}$

$=$

$b_{2}$




$x_{4}$

$+2x_{5}$

$=$

$b_{3}$





$0$

$=$

$b_{4}$


Here too we have solutions iff $b_{4}=0$, and if $b_{4}=0$, we have the freedom to choose the nonpivotal variables $x_{3}$ and $x_{5}$ as we please. But now the formulas for fixing the pivotal variables $x_{1}$, $x_{2}$ and $x_{4}$ in terms of the nonpivotal ones are a bit harder.
Class notes
Scan of Week 11 Lecture 1 notes
Lecture_notes november21st