Confusion about how the determinant changes when all rows are multiplied by a scalar

Question

I am having some trouble thinking about properties of the determinant.

I understand why it is true that if $B$ is the matrix obtained from an $n \times n$ matrix $A$ by multiplying a row by a scalar $k$ then $$\det(B)=k\det(A),$$ but I don't understand why if you multiply every row then you get $$\det(B)=k^{n}\det(A).$$

I am confused because I thought you only do a row expansion along one row or column anyways to calculate the determinant? I mean, how does the expansion take account for this, what is the intuition, etc?

Thank you in advance.

Answer 1

There are a few ways to see this. If you scale each row by $k$ one at a time, you'll pick up a factor of $k$ each time from the formula you have above ($\det B = k\det A$). Geometrically, the determinant is a way to measure volume where the rows (or columns) correspond to the sides of a parallelepiped. If you take a unit cube and scale the whole thing by $k$, then you now have a $k\times\cdots\times k$ cube. The volume of this cube is now $k^n$, whereas your original had volume $1$. A similar argument works for a general parallelepiped.

If you want to appeal to the cofactor expansion, then each cofactor matrix has picked up a factor of $k$. If you continue to break the cofactors down accordingly, you'll end up picking up a factor of $k^n$ multiplying the original determinant. (I'm hiding an induction argument here, but the general idea is there.)

Answer 2

Let $B_1$ the matrix obtained when you take $k$ times the first row of $A$, and for $1\le r\le n$ define $B_r$ as the matrix obtained when the $r$-th row of $B_{r-1}$ is multiplied by $k$, it is easy to see that \begin{align*} \det B_n&=k\det B_{n-1}\\ &=kk\det B_{n-2}\\ &\vdots\\ &=\underbrace{k\ldots k}_{n-1\text{ times}}\det B_1\\ &=k^{n-1}k\det A \end{align*} Observe $B_n=kA$.

Answer 3

If you understand the first part, the second part should be clear as well. You are multiplying the first row by $k$ (which multiplies the determinant by $k$), then the second row by $k$ (multiplying the determinant by $k$ again), etc., until you do $n$ row multiplications which multiplies the determinant by $k^n$.

As for the second part (assuming you are talking about minors), the determinants of the matrices in the expansion also changed. This forms another proof of your statement. Clearly $\det k A = k \det A$ for a $1 \times 1$ matrix. We want to show that for an $n\times n$ matrix $A$ that $\det kA = k^n \det A$. Proceed by induction and assume the statement holds for all matrices with dimensions $< n$. Note that in the minor expansion, there will be a factor of $k$ out front (as you note), but the determinant of all of the minors will also be multiplied by $k^{n-1}$ (induction). Since $k \cdot k^{n-1} = k^n$, your statement follows.