Understand the rank of $ \begin{bmatrix} A&b\\ b^{*}&0 \end{bmatrix}$

Question

Let $A \in M_n(C)$ and $b$ be a column vector of n complex complements. Denote $\widetilde A = \begin{bmatrix} A&b\\ b^{*}&0 \end{bmatrix} $ If $rank(\widetilde A)=rank(A)$, which of the following is true?

(a) $Ax = b$ has infinitely many solutions.

(b) $Ax = b$ has a unique solution.

(c) $\widetilde Ax = 0$ has only solution x = 0.

(d) $\widetilde Ax = 0$ has nonzero solutions.

Zhang, Fuzhen. Linear Algebra

I am assuming that $b^*$ is the conjugate transpose. Could you help me construct an example $A$ and $\widetilde A$ such as $rank(\widetilde A)=rank(A)$ ? I have hard time doing that without assuming that $b=0$.

Answer 1

Here is one example: $$ \tilde A = \begin{pmatrix} -1 & 0 & -1 \\ 0 & 1 & 1 \\ -1 & 1& 0 \end{pmatrix} $$

Answer 2

Hint: If both matrices have the same rank (column rank = row rank), then $b$ lies in the column space of $A$ and so there is a linear combination of the columns of $A$ which gives $b$.