Given $b$ and given $c$, how do you solve for $A$, in $Ab = c$? This might be silly, but let us say that I have a matrix $A$, (a rotation matrix actually), where $A = \begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix} $. I am given $b = \begin{bmatrix} 3 \\ 4\end{bmatrix}$, and $c = \begin{bmatrix} -4.95 \\ -0.705 \end{bmatrix}$
I am somewhat stumped as to how to approach this... this is different than being given $A$ and $c$, and solving for $b$. Anyway, I can solve it geometrically, but how may I solve it algebraically? Thanks.
 A: 2D case:
Let $b=[\beta_1,\beta_2]^T$ and $c=[\gamma_1,\gamma_2]^T$. Obviously, for the existence of a rotation $A$ such that $Ab=c$, we must again have that $\|b\|_2=\|c\|_2$. Let $A$ be of the form $A=\begin{bmatrix}c&-s\\s&c\end{bmatrix}$.
We want $Ab=c$, which is equivalent to
$$
\beta_1 c - \beta_2 s = \gamma_1, \quad \beta_1 s + \beta_2 c = \gamma_2.
$$
Pick your favourite method for solving linear systems (actually, Cramer's rule is pretty neat in this case) and get
$$
c = \frac{\beta_1\gamma_1+\beta_2\gamma_2}{\beta_1^2+\beta_2^2},
\quad
s = \frac{\beta_1\gamma_2-\beta_2\gamma_1}{\beta_1^2+\beta_2^2}.
$$
We should verify that $c^2+s^2=1$:
$$
c^2+s^2=
\frac{(\beta_1\gamma_1 + \beta_2\gamma_2)^2+(\beta_1\gamma_2 - \beta_2\gamma_1)^2}{(\beta_1^2 + \beta_2^2)^2}
=
\frac{\beta_1^2\gamma_1^2+\beta_2^2\gamma_2^2+\beta_1^2\gamma_2^2+\beta_2^2\gamma_1^2}{(\beta_1^2 + \beta_2^2)^2}
=\frac{(\beta_1^2+\beta_2^2)(\gamma_1^2+\gamma_2^2)}{(\beta_1^2 + \beta_2^2)^2}.
$$
We have $c^2+s^2=1$ provided that $\beta_1^2+\beta_2^2=\gamma_1^2+\gamma_2^2$ which is true since $\|b\|_2=\|c\|_2$ by assumption. Now since we have two numbers $c$ and $s$ such that $c^2+s^2=1$, we can find a $\theta\in[0,2\pi)$ such that $c=\cos\theta$ and $s=\sin\theta$.

$n$D case, Householder reflections:
For larger matrices, the easiest way is probably to use Householder reflections.
Again, assume $b$ and $c$ to be nonzero vectors (of the same dimension) such that $\|b\|_2=\|c\|_2$.
A general Householder transformation can be written in the form
$$\tag{HT}
A=I-2\frac{vv^T}{v^Tv}.
$$
Applying $A$ on $b$ we get
$$
Ab=\left(I-2\frac{vv^T}{v^Tv}\right)b=b-2\frac{v^Tb}{v^Tv}v.
$$
Since we want $Ab=c$, the equation above implies that we must have $v\in\mathrm{span}\{b,c\}$. Note that it does not matter how we scale $v$ so let $v=b+\alpha c$ for some scalar $\alpha$.
We have
$$\tag{1}
Ab=\left(1-2\frac{v^Tb}{v^Tv}\right)b-2\alpha\frac{v^Tb}{v^Tv}c.
$$
We want to get $Ab=c$, so the coefficient by $b$ in (1) must be zero. Using the expression for $v$, this is equivalent to
$$\tag{2}
0=1-2\frac{v^Tb}{v^Tv}=1-2\frac{b^Tb+\alpha c^Tb}{b^Tb+2\alpha c^Tb+\alpha^2 c^Tc}.
$$
From $\|b\|_2=\|c\|_2$, we have $b^Tb=c^Tc$ so we set in the equation above $b^Tb=c^Tc=\beta$. Then (2) is true iff
$$
0=1-2\frac{\beta+\alpha c^Tb}{(1+\alpha^2)\beta+2\alpha c^Tb} \quad\Leftrightarrow \quad
\alpha^2=1.
$$
Hence we can choose $\alpha$ to be $\pm 1$.
For any such $\alpha$ (we have $v^Tb/v^Tv=1/2$), we obtain
$$
Ab=-2\alpha\frac{v^Tb}{v^Tv}c=-\alpha c.
$$ 
It follows that with $\alpha=-1$, we get $Ab=c$. Of course, we could also take $\alpha=1$ and multiply the resulting $A$ with $-1$ to get $Ab=c$ again.

The Householder transformations which map a nonzero $b$ to $c$ (of equal 2-norms) are given by
  \begin{align}
A_1&=I-2\frac{v_1v_1^T}{v_1^Tv_1}, & v_1&=b-c, \\
A_2&=2\frac{v_2v_2^T}{v_2^Tv_2}-I, & v_2&=b+c.
\end{align}


$n$D case, Givens rotations:
If you really insist of $A$ being a rotation, what I mean is a product of elementary rotations (similar to the 2D case), you can proceed as follows (maybe there's another way, this is just the simplest approach I can think of now):


*

*There is a standard procedure which uses Givens rotations to transform a nonzero vector $b$ to $\rho e_1$, where $\rho=\|b\|_2=\|c\|_2$ and $e_1=[1,0,\ldots,0]^T$. In this procedure, a sequence of Givens rotations $G_2^{(b)},\ldots,G_n^{(b)}$ is constructed such that $G^{(b)}b=\rho e_1$, where $G^{(b)}=G_2^{(b)}\cdots G_n^{(b)}$ is the product of these elementary rotations ($G_n^{(b)}$ annihilates the $n$-th entry of $b$, $G_{n-1}^{(b)}$ annihilates the $(n-1)$-st entry of $G_n^{(b)}b$,..., $G_2^{(b)}$ annihilates the second entry of $G_3^{(b)}\cdots G_n^{(b)}$b).

*Similarly, construct a sequence of elementary Givens rotations such that their product $G^{(c)}$ applied to $c$ makes it equal to $\rho e_1$ (this is OK, since the norms of $b$ and $c$ are equal).

*We have hence that both $G^{(b)}b=\rho e_1$ and $G^{(c)}c=\rho e_1$, so $G^{(b)}b=G^{(c)}c$ and therefore $Ab=c$ with $A=(G^{(c)})^TG^{(b)}$.
Note that $A$ is a product of elementary rotations so that it is a rotation itself (you can consider it as successively rotating $b$ along the axes of the $n$-dimensional Euclidean space). 
This approach is somewhat less straightforward and slightly harder to implement (due to the use of the "standard procedure") than the previous one based on Householder reflections which leads to just a simple formula for $A$. Let me know if there'd be something unclear (like, e.g., that "standard procedure") :-)
A: If ${\bf b}$ is rotated to ${\bf c}$ in ${\Bbb R}^3$ then the axis of rotation will be in the direction of ${\bf a}={\bf b}\times{\bf c}$, so now all we have to do is find the angle.  This can be done by applying the dot product formula to ${\bf b}$ and ${\bf c}$.
Now choose an orthonormal basis for ${\Bbb R}^3$ including the unit vectors $\hat{\bf a}$ and $\hat{\bf b}$.  The matrix of the rotation with respect to this basis will be
$$R=\pmatrix{1&0&0\cr 0&\cos\theta&-\sin\theta\cr 0&\sin\theta&\cos\theta\cr}\ ,$$
and this can be converted by the usual procedure to a matrix with respect to the standard basis.
A: One way to solve a system of equations is to reduce the matrix to triangular. Well, take a triangular $A$. That gives you a system of equations that is easy to solve row by row.
