Determinant properties of a 4x4 matrix

Question

Let $A$ be a $4x4$ matrix with $\det(A) = -3$.

How would you solve $$\det((2(2A^T)^{-1}))^T$$

1. I know that $\det(A)=\det(A^T)$ = $-3$

2. I believe that $\det(2A) = 2^{k}\det(A)$ (in this case, $k$ being 4) = $-48$

3. I believe that $$\det(2(2A^T)^{-1}) = \frac{2}{\det(2A^T)}$$

So... under these circumstances I got $\frac{-1}{24}$, which is the wrong answer as apparently it is wrong(answer is $-1/3$

Answer 1

You should find $$ \det((2(2A^T)^{-1}))^T = \det(2(2A^T)^{-1}) = 2^4 \det[(2A^T)^{-1}] =\\ \frac{2^4}{\det(2A^T)} = \frac{2^4}{2^4 \det(A^T)} = \frac{2^4}{2^4 \det(A)} = -\frac 1{3} $$

Answer 2

$\det((2(2A^T)^{-1})^T) = \det((2(2A)^{-1})^T) = \det((2(2A)^{-1}) = \det(A^{-1}) = \displaystyle \frac{1}{\det(A)}$

Answer 3

You have little mistake.

We have,

$$det(2(2A^T)^{-1}) = \frac{2^4}{det(2A^T)}$$

Answer 4

Hints:

Since $\;\det A^t=\det A\;$ , we get $\;\det\left(2(2A^t)^{-1}\right)^t=\det\left(2(2A^t)^{-1}\right)\;$

Also, we know that for any scalar $\;k\;,\;\;\det(kA)=k^n\det A\;,\;\;n=$ the order of the (square, of course) matrix $\;A\;$ .

Finally, the product theorem tells us that for a regular matrix $\;A\;$ we have $\;\det(A^{-1})=\frac1{\det A}\;$

Well, now do the calculations...