Prove (or disprove) property of determinant: $\;\det(qA) = q^{n} \det(A).$

Question

Let $A$ be a square matrix. Prove (or disprove) the following: $$\det(qA) = q^{n} \det(A).$$

I tried disproving it with counterexamples but I could not find one. Is there a counterexample I'm overlooking?

Answer 1

You couldn't find a counterexample because there is none. (That is, the proposition is true!)

Do you remember how elementary row operations impact the determinant of a matrix?

In particular, if we multiply a row of a matrix $A$ by $q$, we need to multiply the determinant of $A$ by a factor of $q$. Do this $n$ times, once for each of $n$ rows, and you end with $$\det (qA) = q^n \det A$$

Answer 2

Let $A$ be $n \times n$. Then it's true. Consider such a matrix $C$ that $c_{ii}=q$ for $i=1,2,\cdots,n$ and $c_{ij}=0$ if $i \neq j$. It's clear that $\det C=q^n$ Then:

$$qA=CA$$

Next:

$\det qA= \det CA = \det C \cdot \det A=q^n \det A$.

Answer 3

Just rewrite $qA$ as $qE_nA$ , where $E_n$ is the unit-matrix.

Then we obtain: $\det(qA)=\det(qE_nA)=\det(qE_n)\det(A)=q^n\det(A)$