TITLE: Prove $\det(kA)=k^n\det A$
QUESTION [9 upvotes]: Let $A$ be an $n \times n$ matrix and $k$ be a scalar. Prove that $\det(kA)=k^n\det A$.

I really don't know where to start. Can someone give me a hint for this proof?

REPLY [14 votes]: First, let's recall what multiplication of a matrix by a scalar means:
When we multiply $\;kA$, where $A$ is an $n\times n$ matrix, and $k$ a scalar, then 
every entry $a_{ij}$ of matrix $A$ is multiplied by $k$: i.e. $a_{ij}\mapsto ka_{ij}$ for each $a_{ij}$. That means for each row $i,\;0\leq i\leq n,\;$ we can factor out $k$.
Now...Recall the elementary row operations you've learned, and how each one of them affects the determinant of the matrix on which it is operating. Specifically, when any one row is multiplied by the scalar $k$, the determinant of $A$ becomes $\;k\det A$. So given that scalar multiplication of a square $n\times n$ matrix is equivalent to "row operating" on $n$ rows (by multiplying each row by the scalar $k$), we can conclude that $$\large\det (kA) = \underbrace{k\cdot k\cdot\cdots \cdot k}_{\large n \;\text{times}}\det A = k^n\det A$$

REPLY [2 votes]: Just another way of doing it:
If $\lambda_1,\lambda_2,\cdots,\lambda_n$ are the eigenvalues of a matrix, then $\det(A)=\lambda_1\times\lambda_2\times\cdots\times\lambda_n$.
Let $\lambda$ be any eigenvalue of $A$ and $x$ be its corresponding eigenvector. Now, $Ax=\lambda x$, then it is easy to see that $k\lambda$ is the eigenvalue of $kA$ with the same eigenvector. Now, using the fact that each eigenvalue is being multiplied by $k$ and the determinant is product of eigenvalues, your proof follows.
I do wish to point out that the only reason I put this proof in is for fun. It is just another way of doing it. What you should be doing ideally is to work from the definition of a determinant taught to you.