Tangent plane at a point to surface $\mathbf{F}=\mathbf{0}$ Let $\mathbf{F}:\mathbb{R}^5\to\mathbb{R}^3, \mathbf{F}\begin{pmatrix}x_1\\x_2\\y_1\\y_2\\y_3\end{pmatrix}=\begin{bmatrix}2x_1+x_2+y_1+y_3-1\\x_1x_2^3+x_1y_1+x_2^2y_2^3-y_2y_3\\x_2y_1y_3+x_1y_1^2+y_2y_3^2\end{bmatrix}$ and $\mathbf{a}=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}$.
The equation $\mathbf{F}=\mathbf{0}$ defines $\mathbf{y}=\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}$ as a function of $\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}$ near $\mathbf{a}$ since $\mathbf{F(a)}=\mathbf{0}$, $\mathbf{F}$ is $\mathcal{C}^1$ and $\frac{\partial\mathbf{F}}{\partial\mathbf{y}}(\mathbf{a})=\begin{bmatrix}1 & 0 & 1\\0 & 1 & -1\\ 1 & 1 &1 \end{bmatrix}$ is nonsingular. So we have that near $\mathbf{a}$ there is a $\mathcal{C}^1$ function $\mathbf{\phi}$ such that $\mathbf{y}=\mathbf{\phi(x)}$ and $D\mathbf{\phi}\begin{pmatrix}0\\ 1\end{pmatrix}=-\left(\frac{\partial\mathbf{F}}{\partial\mathbf{y}}(\mathbf{a})\right)^{-1} \left(\frac{\partial\mathbf{F}}{\partial\mathbf{x}}(\mathbf{a})\right)=\begin{bmatrix}-3 & -5\\1 & 2\\1 & 4\end{bmatrix}$.
My question: is it correct to say that the tangent plane at $\mathbf{a}$ of the surface $\mathbf{F}=\mathbf{0}$ is the graph of
$\mathbf{g}:\mathbb{R}^2\to\mathbb{R}^5$
$\mathbf{g}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+x_1\begin{bmatrix}1\\0\\\frac{\partial\phi}{\partial x_1}\begin{pmatrix}0\\1\end{pmatrix}\end{bmatrix}+x_2\begin{bmatrix}0\\1\\\frac{\partial\phi}{\partial x_2}\begin{pmatrix}0\\1\end{pmatrix}\end{bmatrix}=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+x_1\begin{bmatrix}1\\0\\-3\\1\\1\end{bmatrix}+x_2\begin{bmatrix}0\\1\\-5\\2\\4\end{bmatrix}?$
 A: The result is correct. Another way to check this is that the tangent plane at the point can be obtained by translating the subspace $\mathbf{N}([D\mathbf{F(a)}])$ so that it passes through $\mathbf{a}$.
A basis of $N(D\mathbf{F}(\mathbf{a}))$ is $\left\{ 
  \begin{bmatrix}4\\-1\\-7\\2\\0\end{bmatrix}
  , \begin{bmatrix}-2\\1\\1\\0\\2\end{bmatrix}
   \right\}$ so the tangent plane at $\mathbf{a}$ can be written as the graph of: $\mathbf{f}:\mathbb{R}^2\to\mathbb{R}^5,\ \mathbf{f}(s,t)=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+s\begin{bmatrix}4\\-1\\-7\\2\\0\end{bmatrix}+t\begin{bmatrix}-2\\1\\1\\0\\2\end{bmatrix}$ which is equivalent to the graph of $\mathbf{g}:\mathbb{R}^2\to\mathbb{R}^5,\ \mathbf{g}(s,t)=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+s\begin{bmatrix}1 \\ 0 \\ -3 \\ 1 \\ 1\end{bmatrix}+t\begin{bmatrix}0 \\ 1 \\ -5 \\ 2 \\ 4\end{bmatrix}$ (which is my solution above) since $\left\{ 
  \begin{bmatrix}1 \\ 0 \\ -3 \\ 1 \\ 1\end{bmatrix}
  , \begin{bmatrix}0 \\ 1 \\ -5 \\ 2 \\ 4\end{bmatrix}
   \right\}$ is also a basis for $\mathbf{N}([D\mathbf{F(a)}])$.
Note that the tangent plane is in $\mathbb{R}^5$ because it must be tangent to the the graph of $\mathbf{\phi}:\mathbb{R}^2\to\mathbb{R}^3$ which parametrizes the surface $\mathbf{F}=\mathbf{0}$ near $\mathbf{a}$ and which is in $\mathbb{R}^5$.
(Thanks to user Ted Shifrin for the help)
