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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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    @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?2012-06-03

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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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    Oh, he deleted it in the meantime - probably because he saw that you already had posted an answer. What I meant was that $A^3B^2-A^2B^3=A^2(A-B)B^2$.2012-06-03