06-240/Classnotes For Tuesday November 21

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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_1x=b or A_2x=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 non-pivotal 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 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.

Class notes

Scan of Week 11 Lecture 1 notes

Lecture_notes november21st