06-240/Classnotes For Tuesday November 21

06-240/Navigation Panel   # Week of... Notes and Links 1 Sep 11 About, Tue, HW1, Putnam, Thu 2 Sep 18 Tue, HW2, Thu 3 Sep 25 Tue, HW3, Photo, Thu 4 Oct 2 Tue, HW4, Thu 5 Oct 9 Tue, HW5, Thu 6 Oct 16 Why?, Iso, Tue, Thu 7 Oct 23 Term Test, Thu (double) 8 Oct 30 Tue, HW6, Thu 9 Nov 6 Tue, HW7, Thu 10 Nov 13 Tue, HW8, Thu 11 Nov 20 Tue, HW9, Thu 12 Nov 27 Tue, HW10, Thu 13 Dec 4 On the final, Tue, Thu F Dec 11 Final: Dec 13 2-5PM at BN3, Exam Forum Register of Good Deeds Add your name / see who's in! edit the panel

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 $A_{1}x=b$ or $A_{2}x=b$. 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.