Matrices.

**scofield131** · May 26th 2010, 10:37 AM

(Sorry, posting a lot of threads tonight. Exams coming up, past papers are surprisingly unlike my notes

)

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

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

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

**Plato** · May 26th 2010, 10:51 AM

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

**scofield131** · May 26th 2010, 11:14 AM

Originally Posted by Plato

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

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?

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

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

So, AB - BA :

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

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?

**Plato** · May 26th 2010, 11:59 AM

If you have any hope of passing this test then you better improve basic skills.
$BA = \left( {\begin{array}{*{20}c} x & w \\ z & y \\ \end{array} } \right) $

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