Matrices.

Nov 2009
25
0
(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?





 

Plato

MHF Helper
Aug 2006
22,455
8,631
Let \(\displaystyle B = \left( {\begin{array}{*{20}c}
w & x \\ y & z \\ \end{array} } \right)\)
Find \(\displaystyle w,~x,~y,~\&~z\) so that \(\displaystyle AB - BA = \left( {\begin{array}{*{20}c} 1 & 0 \\ 0 & { - 1} \\
\end{array} } \right)\)
 
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Reactions: scofield131
Nov 2009
25
0
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?
 

Plato

MHF Helper
Aug 2006
22,455
8,631
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)
\)
 
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