Determinant of matrix times a constant.

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.

Solution 1:

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.

Solution 2:

To elaborate on the above answer, the formula for the determinant is $$\operatorname{det}(A)=\sum_{\sigma\in S_n}sign(\sigma)\Pi_{i=1}^na_{i,\sigma_i}$$ so $$\operatorname{det}(kA)=\sum_{\sigma\in S_n}sign(\sigma)\Pi_{i=1}^nka_{i,\sigma_i}$$ $$=k^n\operatorname{det}(A)$$

Solution 3:

Note that the matrix $kA$ has elements $[kA]_{ij}=kA_{ij}$, where $A_{ij}$ are the elements of $A$. If we were to calculate the determinant expression formula, each term has the factor $k$ appearing $n$ times, where $n$ is the dimension of the matrix. You can factor these out from the entire expression, and you're left with something proportional to $\det A$.

Example:

\begin{align} \det \left(k\begin{bmatrix} A_{11} & A_{12}\\ A_{21} & A_{22}\\ \end{bmatrix}\right) =& \det \begin{bmatrix} kA_{11} & kA_{12}\\ kA_{21} & kA_{22}\\ \end{bmatrix}\\ = & kA_{11}kA_{22}-kA_{21}kA_{12}\\ = & k^2(A_{11}A_{22}-A_{21}A_{12})\\ =& k^2\det A \end{align}