# Matrices.

#### scofield131

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

• scofield131

#### scofield131

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
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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