Ratio of area formed by transformed and original sides of a parallelogram

Question

I am interested in finding the ratio of area formed by transformed and original sides of a parallelogram, given by: $$\frac{\|Ma\times Mb\| }{\| a\times b \|}$$ $M$ is a $3 \times 3$ matrix and $ a, b$ are vectors with 3 components each ($a,b$ are sides of the original parallelogram and $Ma, Mb$ are sides after transformation). I was thinking of using the identity: $$(M a)\times (M b)=\det(M) (M^{\rm T})^{-1}(a\times b)$$

which is proved here but its not working out. Is there any way to find this ratio?

EDIT

Motivation: we can write the ratio of volumes of parallelepipeds in terms of only $\det (M)$ i.e. $$\frac{\|Ma.(Mb\times Mc)\|}{\|a.(b\times c)\|}=\det(M)$$ so I was thinking whether a similar expression in terms of only $M$ was possible.

Answer 1

The ratio in general doesn't just depend on $M$, it depends on $a$ and $b$ too.

For example, for $$M = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ Then, for $a = {\bf i}$, $b = {\bf j}$, the ratio is $$\frac{||M{\bf i} \times M{\bf j}||}{{\bf i} \times {\bf j}} = \frac{||{\bf i} \times {\bf j}||}{||{\bf i} \times {\bf j}||} = 1 ,$$ but for $a = {\bf i}$, $b = {\bf k}$, the ratio is $$\frac{||M{\bf i} \times M{\bf k}||}{{\bf i} \times {\bf j}} = \frac{||{\bf i} \times {\bf 0}||}{||{\bf i} \times {\bf j}||} = \frac{||\bf 0||}{||{\bf k}||} = \frac{0}{1} = 0 .$$

Of course, in the special case $M = \lambda I$, the ratio is $\lambda^2$, which is independent of $a, b$.

Answer 2

Following this formula the ratio can be written as $$\sqrt\frac{\det((MA)^TMA)}{\det(A^TA)}=\sqrt\frac{\det (A^T(M^TM)A)}{\det(A^TA)}$$ where $A=\pmatrix{a & b}$.