If I have to vectors x and b, how can I solve matrix A in Ax = b? Or is this not possible?
Eg:
x, y, x' and y' are known. Is it possible to get a0-a3 and b0-b3?Code:Ax = b [a0, a1, a2, a3][x] = [x'] [b0, b1, b2, b3][y] [y']
Thanks
If I have to vectors x and b, how can I solve matrix A in Ax = b? Or is this not possible?
Eg:
x, y, x' and y' are known. Is it possible to get a0-a3 and b0-b3?Code:Ax = b [a0, a1, a2, a3][x] = [x'] [b0, b1, b2, b3][y] [y']
Thanks
$\displaystyle \displaystyle \begin{align*}\mathbf{Ax} &= \mathbf{b} \\ \mathbf{A}^T\mathbf{Ax} &= \mathbf{A}^T\mathbf{b} \\ (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{Ax} &= (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{b} \\ \mathbf{x}&= (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{b}\end{align*}$
I have also realised that your matrix equation as written makes no sense, as the dimensions of the matrices $\displaystyle \displaystyle A$ and $\displaystyle \displaystyle x$ don't allow multiplication. Are you sure the equation wasn't
$\displaystyle \displaystyle \left[\begin{matrix}a_0 & b_0 \\ a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3\end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}c_0 \\ c_1 \\ c_2\\ c_3\end{matrix}\right]$
because that WILL work...
Use a similar method - postmultiply both sides by $\displaystyle \displaystyle \mathbf{x}^T$ in order to create a square matrix, then postmultiply both sides by $\displaystyle \displaystyle (\mathbf{xx}^T)^{-1}$ in order to get rid of the $\displaystyle \displaystyle x$ terms.