The determinant of matrix product follows the following property unlike it's sum. $$ \det (AB) = \det A\det B $$ Why is this true? What's the easiest proof?
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2Once you've studied elementary row operations, the easiest way is probably based on the effect of multiplying by corresponding elementary matrices on the determinant (of a square matrix). – hardmath Dec 14 '16 at 23:41
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'unlike it's sum'? – StubbornAtom Dec 15 '16 at 11:39
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One definition of the determinant of an $n\times n$ matrix $M$ is that it is the only $n$-linear alternating form on $M_n(K)$ which takes the value $1$ on $I_n$.
Now the map $\;\begin{aligned}[t] M_n(K)&\longrightarrow K\\ M&\longmapsto\det(AM)\end{aligned}$ is an $n$-linear alternating form w.r.t. the columns of $M$. Thus it is equal to $\lambda \det M$ for some $\lambda$. If you take $M=I_n$, you get $$\det(AI_n)=\lambda \det I_n=\det A, \quad\text{so}\quad\det(AM)=\det A\cdot\det M.$$
Bernard
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