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How to calculate determinant of this matrix?

$\left[\begin{array}{cc} 1 & 2 \\ 0 & -2 \\ \end{array}\right]^3 . \left[\begin{array}{cc} 2 & 3 \\ -1 & 1 \\ \end{array}\right]^2-\left[\begin{array}{cc} 1 & 2 \\ 0 & -2 \\ \end{array}\right]^2 . \left[\begin{array}{cc} 2 & 3 \\ -1 & 1 \\ \end{array}\right]^3$

(It's an exam question, so please explain it.) Thanks..

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What do you mean by "It's an exam question, so please explain it."? Is this an exam you are taking? If it were not an exam question, would you not want an explanation? –  Jonas Meyer Jun 3 '12 at 17:41
    
@JonasMeyer: No, I mean I need a good explanation. Not just a number. –  jiun Jun 3 '12 at 17:48
    
@DavidMitra: Thanks, it was pointless –  jiun Jun 3 '12 at 17:49
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Do you know how to add/subtract and multiply matrices? If so, then you should also know how to square and cube them. This should be sufficient to evaluate the expression to a single $2\times2$ matrix, and you can take the determinant of that (the formula $ad-bc$ is relatively easy to memorize). –  anon Jun 3 '12 at 17:51
    
@anon: Yes, I know, but should I first cube and square them, and subtract the result and then calculate the determinate? there isn't any formula? –  jiun Jun 3 '12 at 17:55

1 Answer 1

up vote 0 down vote accepted

Consider the first matrix as $A$ and the second one as $B$ . Then you have the form

$A^3B^2-A^2B^3$ which is equal to $A^2B^2(A-B)$ Once you know $A-B$ its all in the form of products . You can find the determinant easily.

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Ananda: Be careful - matrix multiplication is not commutative in general. (See David Mitra's answer.)... But you found the right trick for the problem. You even get the correct result this way, since $|XY|=|X||Y|=|Y||X|=|YX|$. –  Martin Sleziak Jun 3 '12 at 17:57
    
@Martin Sleziak : Very true . By the way where is David Mitra's answer ? –  Theorem Jun 3 '12 at 18:00
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Oh, he deleted it in the meantime - probably because he saw that you already had posted an answer. What I meant was that $A^3B^2-A^2B^3=A^2(A-B)B^2$. –  Martin Sleziak Jun 3 '12 at 18:02

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