Evaluating the determinants of matrices A and B I am stuck on a homework question that asks to evaluate the det(A) and det(B). We are only given the following information: A and B are 3x3 matrices, such that det((2A)⁻¹Bᵀ)=1 and det(4B⁻¹A³)=1/2, how could I solve this?
 A: $\det (AB) = \det (A)\det(B)\\
\det (A^n) = \det (A)^n\\
\det (A^{-1}) = \frac 1{\det(A)}$
Transposition does not change determinants
Scalar multiplication multiplies every element in $A$ by that scalar.
When we take the determinant that scalar gets multiplied repeatedly equal to the number of rows in the matrix.
$\det (cA) = c^n\det(A)$
let $a = \det A, b = \det B$
$\det ((2A)^{-1}B^T) = \frac b{8a} = 1$
You can work out the other one, and then solve for $a, b$
A: I'm going to work this out for $n \times n$ matrices, since there is really no more effort involved.
The solution depends almost entirely on the three properties
$\det(XY) = (\det X)(\det Y), \tag{1}$
and
$\det X^T = \det X, \tag{2}$
which hold for any two $n \times n$ matrices $X$, $Y$,and the fact that $\det D$, where $D$ is a diagonal matrix, is the product of its diagonal entries.
From (1), we have, for invertible $X$,
$(\det X)(\det X^{-1}) = \det (XX^{-1}) = \det(I) = 1, \tag{3}$
whence
$\det X^{-1} = (\det X)^{-1}; \tag{4}$
Since
$\det ((2A)^{-1} B^T) = 1, \tag{5}$
we have
$\det (2^{-1}I) \det A^{-1} \det B^T = \det ((2A)^{-1} B^T) = 1, \tag{6}$, 
and also,
$\det (2^{-1}I) = 2^{-n}, \tag{7}$
so using (2) and (4) we find
$2^{-n} (\det A)^{-1} \det B = 1, \tag{8}$
or
$\det B = 2^n \det A; \tag{9}$
also,
$\det (4I) \det B^{-1} \det A^3 = \det (4IB^{-1}A^3) = \det (4B^{-1}A^3) = 2^{-1}, \tag{10}$
and
$\det(4I) = 4^n = 2^{2n}, \tag{11}$
so with the aid of (1), (2) and (4) again
$2^{2n} (\det B)^{-1} (\det A)^3 = 2^{-1}, \tag{12}$
so
$\det B = 2^{2n + 1} (\det A)^3; \tag{13}$
combining (9) and (13):
$2^n \det A = 2^{2n + 1} (\det A)^3; \tag{14}$
thus since we have been (tacitly) assuming $\det A \ne 0$,
$(\det A)^2 = 2^{-(n + 1)}, \tag{15}$
or
$\det A = \sqrt{2^{-(n + 1)}}; \tag{16}$
by (9),
$\det B = 2^n\sqrt{2^{-(n + 1)}} = \sqrt{2^{(n - 1)}}. \tag{17}$
With $n = 3$ we have
$\det A = \sqrt{2^{-4}} = 2^{-2}, \tag{18}$
and
$\det B = \sqrt {2^2} = 2. \tag{19}$
These results are easily checked.
