Find the volume in $\mathbb R^4$ of the set of $x$ with $x^tAx \le 1$. You may use the fact that the volume in $\mathbb R^4$ of the set of $x$ with $|x|^2 = x^tx\le 1$ is $\frac{\pi^2}{2}$.
My work so far:
This is part (b) of the question. From part (a), I was given a symmetric matrix $A$, and I factored $A$ into $LL^t$, using the Cholesky factorization. This factorization reveals easily that $$\det(A) = \det(L)\det(L^t) = (2)(2) = 4.$$
I'm guessing that I have to compute the determinant of some matrix to find the volume of the parallelipiped spanned by the column vectors of that matrix. But I don't know which matrix that would be.
And, I have that the set of x that we need to compute the volume of must satisfy $$\begin{align} x^tAx &= x^t(LL^t)^tx \\ &= ((LL^t)x)^tx \\ &= ((LL^t)^tx)^tx \\ &= ((LL^tx)^tx \\ &= (L^tx)^t(L^tx) \\ &=||L^t(x)||^2 \\ &\le 1 \end{align}$$
Where can I go from here?
So you have a map from $\mathbb R^4 \to \mathbb R^4$; $x \mapsto L^t x$, which maps the set you are interested in to the unit sphere. Since multiplication by a matrix $M$ causes sets to be mapped to new sets whose volume is $\det(M)$ times as large as its original volume, you deduce that the volume of your set is $\frac{\pi^2}2 \div \text{det}(L^t) = \frac{\pi^2}4$.