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If I am given two $2\times2$ polynomial matrices and I need to divide them, what are the steps I need to follow?

I know I need to do right division and left division, and that the answer will have the right quotient and remainder and the left quotient and remainder.

Please show me how to determine these values with an example. I don't want to put up the exact question I need to answer, I'd like to do that myself. I just want an example please.

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If you're in fact evaluating the rational function $\frac{p(x)}{s(x)}$, $p$ and $q$ polynomial, at a matrix argument, it might be a good idea if you take the quotient-remainder decomposition $q(x)+\frac{r(x)}{s(x)}$ first before evaluating at the matrix argument. –  J. M. Jul 31 '11 at 11:51

1 Answer 1

Dividing matrices is multiplying by the inverse, i.e if you want to divide $A$ by $B$ then you calculate $A\cdot B^{-1}$ (right) or $B^{-1}\cdot A$ (left). In order to find the inverse you use the Gauss reduction or the adjoint formula. For $2\times2$ matrices you have: $$\left( \begin{array}{cc} a&b \\ c&d \end{array}\right)^{-1} = \frac{1}{ad-bc}\left( \begin{array}{cc} d&-b \\ -c&a \end{array}\right)$$ It doesn't matter whether the entries are numbers or polynomials.
To elborate on the comments: For example, suppose $A=\left( \begin{array}{cc} x^2&x+1 \\ x+2&x^2+1 \end{array}\right)$ and $B=\left( \begin{array}{cc} x&x \\ x+1&2x \end{array}\right)$. Then you have $B^{-1}= \frac{1}{2x^2-x^2-x}\left( \begin{array}{cc} 2x&-x \\ -x-1&x \end{array}\right)$. Suppose you want to calculate the right quotient. $$AB^{-1}=\frac{1}{x^2-x}\left( \begin{array}{cc} 2x^3-(x+1)^2&-x^3+x^2+x\\ 2x^2+4x-(x+1)(x^2+1)&x^3+x-x^2-2x \end{array}\right)$$ Now divide each entry by the determinant, to get: $$AB^{-1}=\left( \begin{array}{cc} 2x+1&-x \\ -x&x \end{array}\right)+\frac{1}{x^2-x}\left( \begin{array}{cc} -x-1&x \\ 3x-1&-x \end{array}\right)$$ Multiplying both sides by $B$, you get: $$A=\left( \begin{array}{cc} 2x+1&-x \\ -x&x \end{array}\right)\cdot B+\left( \begin{array}{cc} 0&1 \\ 2&1 \end{array}\right)$$ So the right quotient is $\left( \begin{array}{cc} 2x+1&-x \\ -x&x \end{array}\right)$ and the right reminder is $\left( \begin{array}{cc} 0&1 \\ 2&1 \end{array}\right)$.

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It does matter if the entries of the matrix don't come from a field! –  Zhen Lin Jul 31 '11 at 9:39
    
That's right, for that he needs to divide each entree of the resulted matrix by the determinant to get the quotient and the reminder. –  Dennis Gulko Jul 31 '11 at 9:41
    
it's "entry", not the main course :) (even more confusing both from "entrer"--to enter, at least in america entree has come to mean the main course, not the first course...) –  yoyo Jul 31 '11 at 13:41

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