Question

(Least squares) We want to adjust the plane: y=C+Dt+Ez at these four points:
y=3,t=1,z=1
y=5,t=2,z=1
y=6,t=0,z=3
y=0,t=0,z=0.
How do I find C,D,E?

Answer 1

You have a system given in the form
\[\mathbf{Ax}=\mathbf{b}\iff \begin{bmatrix}1&1&1\\1&2&1\\1&0&3\\0&0&0\end{bmatrix}\begin{bmatrix}C\\D\\E\end{bmatrix}=\begin{bmatrix}3\\5\\6\\0\end{bmatrix}\]which has no solution because \(\mathbf{b}\) is not in the column space of \(\mathbf{A}\).

To find the "best" solution, multiply both sides by \(\mathbf{A}^\intercal\):
\[\mathbf{A}^\intercal\mathbf{Ax}=\mathbf{A}^\intercal\mathbf{b}\iff \begin{bmatrix}4&3&5\\3&5&3\\5&3&11\end{bmatrix}\begin{bmatrix}C\\D\\E\end{bmatrix}=\begin{bmatrix}14\\13\\26\end{bmatrix}\]then multiply both sides by the inverse of \(\mathbf{A}^\intercal\mathbf{A}\) and find the "best" \(C,D,E\):
\[\mathbf{x}=\left(\mathbf{A}^\intercal\mathbf{A}\right)\mathbf{A}^\intercal\mathbf{b}\implies \mathbf{x}=\cdots\]

Answer 2

Note that I'm missing an inverse - the last equation should be
\[\mathbf{x}=\left(\mathbf{A}^\intercal\mathbf{A}\right)^{\color{red}{-1}}\mathbf{A}^\intercal\mathbf{b}\implies\cdots\]