# 06-240/Classnotes For Tuesday November 21

## 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:

 ${\displaystyle A_{1}={\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}}$ ${\displaystyle A_{2}={\begin{pmatrix}1&0&-4&0&-6\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}}$

So let us assume row reduction leads us to the systems ${\displaystyle A_{1}x=b}$ or ${\displaystyle A_{2}x=b}$. What does it tell us about the solutions? Let us start from the second system:

${\displaystyle {\begin{pmatrix}1&0&-4&0&-6\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{pmatrix}}={\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}}$ or
 ${\displaystyle x_{1}}$ ${\displaystyle -4x_{3}}$ ${\displaystyle -6x_{5}}$ ${\displaystyle =}$ ${\displaystyle b_{1}}$ ${\displaystyle x_{2}}$ ${\displaystyle +2x_{3}}$ ${\displaystyle -2x_{5}}$ ${\displaystyle =}$ ${\displaystyle b_{2}}$ ${\displaystyle x_{4}}$ ${\displaystyle +2x_{5}}$ ${\displaystyle =}$ ${\displaystyle b_{3}}$ ${\displaystyle 0}$ ${\displaystyle =}$ ${\displaystyle b_{4}}$

Well, quite clearly if ${\displaystyle b_{4}\neq 0}$ this system has no solutions, but if ${\displaystyle b_{4}=0}$ it has solutions no matter what ${\displaystyle b_{1}}$, ${\displaystyle b_{2}}$ and ${\displaystyle b_{3}}$ are. Finally, for any given values of ${\displaystyle b_{1}}$, ${\displaystyle b_{2}}$ and ${\displaystyle b_{3}}$ we can choose the values of ${\displaystyle x_{3}}$ and ${\displaystyle x_{5}}$ (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: ${\displaystyle x_{1}=b_{1}+4x_{3}+6x_{5}}$, ${\displaystyle x_{2}=b_{2}-2x_{3}+2x_{5}}$ and ${\displaystyle x_{4}=b_{3}-2x_{5}}$.

What about the system corresponding to ${\displaystyle A_{1}}$? It is

${\displaystyle {\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{pmatrix}}={\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}}$ or
 ${\displaystyle x_{1}}$ ${\displaystyle +3x_{2}}$ ${\displaystyle +2x_{3}}$ ${\displaystyle +4x_{4}}$ ${\displaystyle +2x_{5}}$ ${\displaystyle =}$ ${\displaystyle b_{1}}$ ${\displaystyle x_{2}}$ ${\displaystyle +2x_{3}}$ ${\displaystyle +3x_{4}}$ ${\displaystyle +4x_{5}}$ ${\displaystyle =}$ ${\displaystyle b_{2}}$ ${\displaystyle x_{4}}$ ${\displaystyle +2x_{5}}$ ${\displaystyle =}$ ${\displaystyle b_{3}}$ ${\displaystyle 0}$ ${\displaystyle =}$ ${\displaystyle b_{4}}$

Here too we have solutions iff ${\displaystyle b_{4}=0}$, and if ${\displaystyle b_{4}=0}$, we have the freedom to choose the non-pivotal variables ${\displaystyle x_{3}}$ and ${\displaystyle x_{5}}$ as we please. But now the formulas for fixing the pivotal variables ${\displaystyle x_{1}}$, ${\displaystyle x_{2}}$ and ${\displaystyle x_{4}}$ in terms of the non-pivotal ones are a bit harder.