$A$ and $B$ are $3\times3$ matrices. $3A-B^2=0$ and $A^2-4B=0$, find possible determinants for A and B

Question

$$3A-B^2=0$$ $$A^2-4B=0$$ I don't know how to solve that. I've tried solving for some general $3a_{ij}$ and express the element that is deducted but it doesn't get me anywhere...

Answer 1

You have $$3A=B^2,$$hence $$3^3\det A=(\det B)^2.$$ Similarly, $$A^2=4B,$$hence $$(\det A)^2=4^3\det B.$$Can you solve this system on $\det A$ and $\det B$?

edit

The solution; let $a=\det A$ and $b=\det B$, then the system writes $$3^3a=b^2,\quad 4^3b=a^2.$$ Square the first equation:

$$b^4=3^6a^2 = 3^6\cdot 4^3 b.$$ The first case is $a=b=0$, which is trivial; suppose that $a\ne0$ and $b\ne 0$, which gives$$b^3=3^6\cdot 4^3$$ or $b=36$, which implies $a=b^2/27 = 48$.