# Matrices.

• May 26th 2010, 10:37 AM
scofield131
Matrices.
(Sorry, posting a lot of threads tonight. Exams coming up, past papers are surprisingly unlike my notes :( )

A = $\displaystyle \begin{pmatrix}0&1\\1&0\end{pmatrix}$

Determine all 2 x 2 matrices B such that AB-BA = $\displaystyle \begin{pmatrix}1&0\\0&-1\end{pmatrix}$

I really don't know how to even start this one. Could anybody give any pointers?

• May 26th 2010, 10:51 AM
Plato
Let $\displaystyle B = \left( {\begin{array}{*{20}c} w & x \\ y & z \\ \end{array} } \right)$
Find $\displaystyle w,~x,~y,~\&~z$ so that $\displaystyle AB - BA = \left( {\begin{array}{*{20}c} 1 & 0 \\ 0 & { - 1} \\ \end{array} } \right)$
• May 26th 2010, 11:14 AM
scofield131
Quote:

Originally Posted by Plato
Let $\displaystyle B = \left( {\begin{array}{*{20}c} w & x \\ y & z \\ \end{array} } \right)$
Find $\displaystyle w,~x,~y,~\&~z$ so that $\displaystyle AB - BA = \left( {\begin{array}{*{20}c} 1 & 0 \\ 0 & { - 1} \\ \end{array} } \right)$

Ok, so I've tried to work though it that way but I've hit a snag. Perhaps I've multiplied wrongly? Can you see the error?

$\displaystyle AB = \left( {\begin{array}{*{20}c} z & y \\ w & x \\ \end{array} } \right)$

$\displaystyle BA = \left( {\begin{array}{*{20}c} z & y \\ x & w \\ \end{array} } \right)$

So, AB - BA :

$\displaystyle B = \left( {\begin{array}{*{20}c} y-z & z-y \\ w-x & x-w \\ \end{array} } \right)$

Which leaves the situation of y always having to be one bigger than z, but z always having to equal y, which can't be true?
• May 26th 2010, 11:59 AM
Plato
If you have any hope of passing this test then you better improve basic skills.
$\displaystyle BA = \left( {\begin{array}{*{20}c} x & w \\ z & y \\ \end{array} } \right)$