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I've a matrix-equation and I'm trying to list the conditions under which that equation can be true. The equation is of the form: $$\small \begin{pmatrix} -A & B+b \\ A+a & -B \end{pmatrix} \cdot \begin{pmatrix} m_1 \\ m_2 \end{pmatrix} = \begin{pmatrix} -c & D \\ C & -d \end{pmatrix} \cdot \begin{pmatrix} r_1 \\ r_2 \end{pmatrix} $$ All numbers are expected to be integer and positive, the small letters have smaller (a and b) or smaller-or-equal (c and d) absolute values than the resp. capital letters. Also A to D , a to d,r1,r2 are given (or more exactly, functions of two problem-parameters) and I'm studying the restrictions that lay on m1 and m2. Because from the configuration of the problem I can assume, the determinant of the most-left matrix is negative, and I wonder, whether this means, that the determinant of the 2x2-matrix on the rhs must also be negative. This must be an extremely simple question but I don't just find the key.

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The determinants of the matrix on the right is $cd - CD$, so if $0 < c \le C$ and $0 < d \le D$, certainly $cd - CD \le 0$, with equality only if $c=C$ and $d=D$. Now with $c=C$ and $d=D$, the right side of your equation is $$\pmatrix{- c r_1 + d r_2\cr c r_1 - d r_2\cr}$$ so adding the two entries we must have $0 = a m_1 + b m_2$, which can't happen if everything is positive.

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ahh, thanks! I just thought, there would be some general rule when determinants and a right-multiplication are involved, and I should perhaps just know it... In more general cases, the determinants can even have opposite sign? Hmm... –  Gottfried Helms Apr 16 '12 at 21:56
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