TITLE: Determinant of matrix times a constant.
QUESTION [13 upvotes]: Prove that $\det(kA) = k^n \det(A)$ for and ($n \times n$) matrix. 
I have tried looking at this a couple of ways but can't figure out where to start. It's confusing to me since the equation for a determinant is such a weird summation.

REPLY [16 votes]: A mnemonic using more advanced stuff, just to recover it from the few determinant rules I manage to remember is:
$Det(kA)=Det(kI_nA)=Det(kI_n)Det(A)=Det\left( \begin{bmatrix}
k & 0 & ... & 0 \\
0 & k & ... & 0 \\
... & ... & ... & ...\\
0 & 0 & ... & k
\end{bmatrix} \right)  Det(A) = k^n Det(A)$
Where I have used $Det(AB)=Det(A)Det(B)$ and a formula for the determinant of diagonal matrices.