06-240/Classnotes For Tuesday November 21

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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 2-5PM at BN3, Exam Forum
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More about the Wongpak Matrices

In Talk:06-240/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:

$\begin{pmatrix}1&0&-4&0&-6\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = \begin{pmatrix}b_1\\b_2\\b_3\\b_4\end{pmatrix}$

$x 1$		$- 4 x 3$		$- 6 x 5$	$=$	$b 1$
	$x 2$	$+ 2 x 3$		$- 2 x 5$	$=$	$b 2$
			$x 4$	$+ 2 x 5$	$=$	$b 3$
				$0$	$=$	$b 4$

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 non-pivotal ones as follows: $x 1 = b 1 + 4 x 3 + 6 x 5$ , $x 2 = b 2 - 2 x 3 + 2 x 5$ and $x 4 = b 3 - 2 x 5$ .

What about the system corresponding to $A 1$ ? It is

$\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = \begin{pmatrix}b_1\\b_2\\b_3\\b_4\end{pmatrix}$

$x 1$	$+ 3 x 2$	$+ 2 x 3$	$+ 4 x 4$	$+ 2 x 5$	$=$	$b 1$
	$x 2$	$+ 2 x 3$	$+ 3 x 4$	$+ 4 x 5$	$=$	$b 2$
			$x 4$	$+ 2 x 5$	$=$	$b 3$
				$0$	$=$	$b 4$

Here too we have solutions iff $b 4 = 0$ , and if $b 4 = 0$ , we have the freedom to choose the non-pivotal 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 non-pivotal ones are a bit harder.

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Class notes

Scan of Week 11 Lecture 1 notes

Lecture_notes november21st

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06-240/Classnotes For Tuesday November 21

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