# 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:

 $A_1=\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}$ $A_2=\begin{pmatrix}1&0&-4&0&-6\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}$

So let us assume row reduction leads us to the systems A1x = b or A2x = 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
 x1 − 4x3 − 6x5 = b1 x2 + 2x3 − 2x5 = b2 x4 + 2x5 = b3 0 = b4

Well, quite clearly if $b_4\neq 0$ this system has no solutions, but if b4 = 0 it has solutions no matter what b1, b2 and b3 are. Finally, for any given values of b1, b2 and b3 we can choose the values of x3 and x5 (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: x1 = b1 + 4x3 + 6x5, x2 = b2 − 2x3 + 2x5 and x4 = b3 − 2x5.

What about the system corresponding to A1? 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
 x1 + 3x2 + 2x3 + 4x4 + 2x5 = b1 x2 + 2x3 + 3x4 + 4x5 = b2 x4 + 2x5 = b3 0 = b4

Here too we have solutions iff b4 = 0, and if b4 = 0, we have the freedom to choose the non-pivotal variables x3 and x5 as we please. But now the formulas for fixing the pivotal variables x1, x2 and x4 in terms of the non-pivotal ones are a bit harder.