Find the values of parameter a so that matrices $A=\begin{pmatrix} 1 &4-a-a^2 \\ 2 & -1 \end{pmatrix}$ and $B=\begin{pmatrix} -a-1 &3 \\ 3 & -5 \end{pmatrix}$ may represent the same bilinear form in different bases.
As I know, if matrix of bilinear form is nonsingular with respect to some basis, it's also nonsingular in any basis. S0, we have
det(A) and det(B) are nonzero. Is it enough condition?
No, it isn't a sufficient condition.
First of all, $A$ has to be symmetric.
Therefore, $a$ is solution of the quadratic equation $a^2+a-2=0$ giving
$$a=1 \ \ \text{or} \ \ a=-2$$
As $\det(A)=-1-4 < 0$, it is compulsory that $\det(B)<0$ as well.
This is only possible with
As, with this choice of $a$, we have the same (positive) sign for the two upper left entries : $A_{11} >0$ and $B_{11} > 0$, we are sure that $A$ and $B$ are equivalent, because their leading principal minors have the same signs (Sylvester law of inertia).
One can look for matrix $S$ such that $S^TBS=A$ ; one finds, using a computer :
$$S=\begin{pmatrix} \sqrt{70} - 2 \sqrt{13} - 3 \sqrt{910}/14 + 6& \sqrt{13} - 3\\ \sqrt{910}/14 - 2& 1\end{pmatrix}$$ (no comments !)