A and B are $3\times3$ matrices. Given det $A = -4$ and det $B = 2$
Finding the $det(2A^{-1}B$ )
With the order of operations would I invert first, multiply by two then multiply by $B$, which is also two.
$(2/1)(1/-4)(2/1) = 4/-4 = -1$
Does that seem right?
On
If $A$ and $B$ are scalars, then $2A^{-1}B$ is also a scalar, and we can speak of the determinant of the scalar as being itself. Hence $\det(2A^{-1}B) = 2(\det A)^{-1}\det B=2\cdot\left(-\frac{1}{4}\right) \cdot 2=-1.$
If $A$ and $B$ are scalar $n\times n$ matrices, then instead we have $\det(2A^{-1}B) = 2(\det A)^{-1}\det B=2\cdot\left(-\frac{1}{4^n}\right) \cdot 2^n=-\frac{1}{2^{n-1}}.$
If you mean that $A$ and $B$ are matrices with $\det A=-4$ and $\det B=2$, then yes, it's as you say, we have $\det(2A^{-1}B) = 2(\det A)^{-1}\det B=2\cdot\left(-\frac{1}{4}\right) \cdot 2=-1.$
No, it's not right.
If $X$ is a $3\times 3$ matrix, then $$ \det(2X)=2^3\det X $$ If $\det Y\ne0$, then $\det(Y^{-1})=(\det Y)^{-1}$. Finally, $\det(XY)=\det X\cdot\det Y$.
Put the pieces together.