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:

[math]\displaystyle{ 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} }[/math]

[math]\displaystyle{ 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} }[/math]

So let us assume row reduction leads us to the systems [math]\displaystyle{ A_1x=b }[/math] or [math]\displaystyle{ A_2x=b }[/math]. What does it tell us about the solutions? Let us start from the second system:

[math]\displaystyle{ \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} }[/math]

or

[math]\displaystyle{ x_1 }[/math] [math]\displaystyle{ -4x_3 }[/math] [math]\displaystyle{ -6x_5 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_1 }[/math] [math]\displaystyle{ x_2 }[/math] [math]\displaystyle{ +2x_3 }[/math] [math]\displaystyle{ -2x_5 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_2 }[/math] [math]\displaystyle{ x_4 }[/math] [math]\displaystyle{ +2x_5 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_3 }[/math] [math]\displaystyle{ 0 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_4 }[/math]

Well, quite clearly if [math]\displaystyle{ b_4\neq 0 }[/math] this system has no solutions, but if [math]\displaystyle{ b_4=0 }[/math] it has solutions no matter what [math]\displaystyle{ b_1 }[/math], [math]\displaystyle{ b_2 }[/math] and [math]\displaystyle{ b_3 }[/math] are. Finally, for any given values of [math]\displaystyle{ b_1 }[/math], [math]\displaystyle{ b_2 }[/math] and [math]\displaystyle{ b_3 }[/math] we can choose the values of [math]\displaystyle{ x_3 }[/math] and [math]\displaystyle{ x_5 }[/math] (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: [math]\displaystyle{ x_1=b_1+4x_3+6x_5 }[/math], [math]\displaystyle{ x_2=b_2-2x_3+2x_5 }[/math] and [math]\displaystyle{ x_4=b_3-2x_5 }[/math].

What about the system corresponding to [math]\displaystyle{ A_1 }[/math]? It is

[math]\displaystyle{ \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} }[/math]

or

[math]\displaystyle{ x_1 }[/math] [math]\displaystyle{ +3x_2 }[/math] [math]\displaystyle{ +2x_3 }[/math] [math]\displaystyle{ +4x_4 }[/math] [math]\displaystyle{ +2x_5 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_1 }[/math] [math]\displaystyle{ x_2 }[/math] [math]\displaystyle{ +2x_3 }[/math] [math]\displaystyle{ +3x_4 }[/math] [math]\displaystyle{ +4x_5 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_2 }[/math] [math]\displaystyle{ x_4 }[/math] [math]\displaystyle{ +2x_5 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_3 }[/math] [math]\displaystyle{ 0 }[/math] [math]\displaystyle{ = }[/math] [math]\displaystyle{ b_4 }[/math]

Here too we have solutions iff [math]\displaystyle{ b_4=0 }[/math], and if [math]\displaystyle{ b_4=0 }[/math], we have the freedom to choose the non-pivotal variables [math]\displaystyle{ x_3 }[/math] and [math]\displaystyle{ x_5 }[/math] as we please. But now the formulas for fixing the pivotal variables [math]\displaystyle{ x_1 }[/math], [math]\displaystyle{ x_2 }[/math] and [math]\displaystyle{ x_4 }[/math] in terms of the non-pivotal ones are a bit harder.