Is this convexification of non-convex constraint correct?

Question

Original problem setup How to handle equality constraints in this problem?

\begin{equation}\tag{1} \begin{array}{c} \min_{\mathbf{b}} \hspace{4mm} \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} \\ s.t \hspace{5mm} \mathbf{b} \in \mathbb{R}^{4} \\ \hspace{9mm}b_0=1\\ \hspace{15mm}b_3=b_1b_2 \end{array} \end{equation}

where $\mathbf{A}$ is a $4\times 4$ positive semi-definite matrix.

I Reformulated this to

\begin{equation}\tag{2} \begin{array}{c} \min_{\mathbf{b}} \hspace{4mm} \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} + \lambda(\mathbf{b}^{T}_{}\mathbf{Z}^{}_{}\mathbf{b}^{}_{}-1) \end{array} \end{equation}

where

\begin{equation} \mathbf{Z} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix} \end{equation}

My Question:

1. Are (1) and (2) exactly same?
2. Does (2) result in convexification of non-convex constraint $b_3=b_1b_2$
3. If (2) is correct, then do we have a closed-form solution for $\mathbf{b}$

(Can someone please edit to place $\mathbf{b}$ exactly below min)

Answer 1

For (1) and (2) to be the same, you have to specify $\lambda$. But even if you add $\max_\lambda$ after $\min_b$, they are not the same. The term in the Lagrangian sets $b_0^2 -2b_1 b_2 + 2b_3^2 = 1$, which does not guarantee that $b_0=1$ or that $b_3=b_1b_2$.

The formulation is also not convex, because for that you need $A+\lambda Z$ to be positive semidefinite.