MHB Does Commutativity Hold for Matrices A and B with a Specific Matrix C?

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The discussion confirms that if matrices A and B satisfy the conditions AC = CA and BC = CB with matrix C defined as C = [[0, 1], [-1, 0]], then commutativity holds, meaning AB = BA. The proof involves constructing matrices A and B in the forms A = [[a, b], [-b, a]] and B = [[w, x], [-x, w]]. The discussion emphasizes the importance of correctly interpreting the matrix multiplication properties to establish this result.

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If A and B are matrices that AC = AC and BC=CB, where C is a matrix whose first row's entries are 0 1 and the second row's entries are -1 0, then AB=BA.
 
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MathHelpBoardsUser said:
If A and B are matrices that AC = AC and BC=CB, where C is a matrix whose first row's entries are 0 1 and the second row's entries are -1 0, then AB=BA.
Is there a typo? Did you mean AC = CA?

-Dan
 
topsquark said:
Is there a typo? Did you mean AC = CA?

-Dan
Yes. I apologize.
 
Okay, so this is more or less a construction proof. You know that
[math]C = \left ( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right )[/math]

Let
[math]A = \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right )[/math]

and
[math]B = \left ( \begin{matrix} w & x \\ y & z \end{matrix} \right )[/math]

So.
1) Using AC = CA show that
[math]A = \left ( \begin{matrix} a & b \\ -b & a \end{matrix} \right )[/math]

2) Using BC = CB show that
[math]B = \left ( \begin{matrix} w & x \\ -x & w \end{matrix} \right )[/math]

3) Using A and B from 1) and 2) show that AB = BA.

-Dan
 

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