06-240/Classnotes For Tuesday November 21

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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.