Math people!
In here Relationships between $\det(A+B)$ and $A+B$ I could find a formula only for $\det(A+B)$ when $A, B\in\mathbb{R}^{2\times2}$. Can a formula for $\det(A+B+C)$, $A, B, C\in\mathbb{R}^{2\times2}$ reduced in the same or similar way?
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I assume you can just use the formula for $A+B$ twice. It won't necessarily give a nice result, though. – Arthur Sep 06 '15 at 07:15
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1@Arthur I definitely agree, since addition of matrices (with entries in an associative algebraic structure) is associative! Right? – MattAllegro Sep 06 '15 at 08:09
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Somehow, this "Math people!" greeting is reminiscent of "crab people" in South Park ;-D – user1551 Sep 06 '15 at 10:49
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Excuse me if it hurts someone. – Asatur Khurshudyan Sep 06 '15 at 11:38
1 Answers
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The determinant is multilinear in the columns (alternately, rows) of the matrices. Therefore if you express $A$ as $(a_1, a_2)$ (two column vectors, if you like) and $B$ as $(b_1, b_2)$, then $\det (A+B) = \det (a_1 + b_1, a_2 + b_2)$ and now you can break off the terms one at a time:
$$\det (a_1 + b_1, a_2 + b_2) = \color{blue}{\det (a_1, a_2 + b_2)} + \color{red}{\det (b_1, a_2 + b_2)}$$ $$= \color{blue}{\det (a_1 ,a_2) + \det (a_1 , b_2)} + \color{red}{\det (b_1, a_2) + \det (b_1, b_2)}.$$
You can do the same thing with any number of terms, such as $\det(A+B+C)$ and in any dimension. The terms do start to build up, though.
Eric Auld
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