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**Prove It** You should know $\displaystyle \det{(AB)} = \det{(A)}\det{(B)}$.

In your case, you are wanting to find $\displaystyle \det{(cA)}$.

Note that $\displaystyle cA = cI_nA$, which is now the product of two matrices, $\displaystyle cI_n$ and $\displaystyle A$.

So $\displaystyle \det{(cA)} = \det{(cI_nA)}$

$\displaystyle = \det{(cI_n)}\det{(A)}$.

Since $\displaystyle cI_n$ is a diagonal matrix, its determinant is the product of the diagonal entries. In this case they are all $\displaystyle c$, so their product is $\displaystyle c^n$. Thus $\displaystyle \det{(cI_n)} = c^n$, and finally we have...

$\displaystyle \det{(cA)} = c^n\det{(A)}$, which completes the proof.