Which version of adjoint is that? The Hermitian adjoint (conjugate transpose), or the version of the adjoint such that $A(\text{adj}\,A)=(\det A)I\,?$

${{A}^{*}}={{A}^{-1}}\cdot \det \left( A \right)$

I'm reading the first part of what you said as a hypothesis that $A$ is not invertible. It's not going to work without that hypothesis. Let's consider three cases: $n=2,$ or $n\ge 3$ and $\operatorname{rank}A=n-1,$ or $n\ge 3$ and $\operatorname{rank}A\le n-2.$ Suppose $n=2.$ We could fold this into the later work, but it's easy enough to brute-force this. $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $A^*=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$ Then by direct computation, $\det(I+A^*)=1+a+d+(ad-bc)$ but $ad-bc=0$ by hypothesis. So this determinant is nonzero if and only if $a+d\ne-1.$ Suppose $n\ge 3$ and $\operatorname{rank}A\le n-2.$ Then any $n-1$ rows of $A$ are linearly dependent, which implies that any $(n-1)\times(n-1)$ submatrix has dependent rows. Hence, $A^*=0.$ In that case, $\operatorname{trace}A^*=0\ne -1$ and $I+A^*=I$ is invertible. This leads us to the "main" case: $n\ge 3$ and $\operatorname{rank}A= n-1.$ Since $AA^*=0,$ we conclude that every column of $A^*$ lies within the null space of $A.$ But the null space of $A$ was one dimensional. We are forced to conclude that $\operatorname{rank}A^*=1.$ From now on, let $B=A^*$ and let the fact that $\operatorname{rank}B=1$ be the only thing we assume about $B.$ A general statement about rank 1 matrices: $B^2=(\operatorname{trace}B)B.$ (We can get that by writing $B=xy^T$ for column vectors $x,y.$) That means that the algebra generated by $B$ consists only of $\operatorname{span}\{I,B\}.$ If $I+B$ has an inverse, it must lie within the algebra generated by $B,$ and hence in that span. We can quickly determine that the coefficient of $I$ must be $1,$ so any possible inverse of $B$ must be of the form $I+\alpha B.$ Multiply that out: $(I+B)(I+\alpha B)=I+(1+\alpha+\alpha(\operatorname{trace}B))B.$ If this is going to equal $I,$ then we must have $\alpha(1+\operatorname{trace}B)=-1.$ If $\operatorname{trace}B=-1,$ then this has no solution and $I+B$ is not invertible. If $\operatorname{trace}B\ne-1,$ then we can solve for $\alpha=\frac{-1}{1+\operatorname{trace}B}$ and verify that $I+\alpha B$ is an inverse of $I+B.$ I'll note that none of this is field-specific; the argument would work over any field. (Although in characteristic 2, we feel silly writing $-1.$)

Since $A$ is singular all the $k$th order minors of $A^*$ where $2\leq k\leq n$ are zero. On the other hand we know that the characteristic polynomial of an $n\times n$ matrix $C$ is $p(\lambda)=\det(\lambda I-C)=\lambda^n-p_{1}\lambda^{n-1}+p_{2}\lambda^{n-2}-...+(-1)^np_{n}$ where $p_{k}$ is the sum of all $\binom{n}{k}$ diagonal minors of order $k$ in $C$ . Therefore the characteristic polynomial of $A^*$ should be $p(\lambda)=\det(\lambda I-A^*)=\lambda^n-\operatorname{trace}(A^*)\lambda^{n-1}$. We have $\begin{matrix}\det(I+A^*)=(-1)^n\det(-I-A^*)=(-1)^{n}p(-1)=(-1)^n((-1)^{n}-\operatorname{trace}(A^*)(-1)^{n-1})=1+\operatorname{trace}(A^*) \end{matrix}$ so $\operatorname{trace}(A^*)\neq-1$ iff $\det(I+A^*)\neq0$.

I should have just said that if the rank of $B$ is $1,$ then the only nonzero eigenvalue of $B$ is the trace of $B.$ If the trace of $B$ is not $-1,$ then $-1$ is not an eigenvalue and $I+B$ is invertible. That's simpler than the last part of my argument above. (In this particular instance, $B=A^*$.) babylon5 reached that point as well, by actually naming the characteristic polynomial of $A^*.$

If we know this lemma ,we can finish this problem quickly Lemma For any $n\times n$ matrix we have$(1)$ if $rankA=n$ then $rankA^{*}=n$ $(2)$ if $rankA=n-1$ then $rankA^{*}=1$ $(3)$ if $rankA< {n-1}$ then $rankA^{*}=0$ and we only need to discuss the case that $rankA=n-1$ with $rankA^{*}=1$ and then we can find that $det(\lambda I-(I+A^{*}))=(\lambda -1)^{n}-tr(A^{*})(\lambda -1)^{n-1}$ and get the proof.

You can also consider the fact that $A^*$ has as eigenvalues the products of $(n-1)$ different eigenvalues of $A$. Since $A$ is non-invertible, it has a null eigenvalue and so $A^*$ has at least $(n-1)$ null eigenvalues. If $A$ has a second null eigenvalue then $A^*$ has only null eigenvalues so $tr({{A}^{*}}) = 0 \ne -1$ and the eigenvalues of ${{I}_{n}}+{{A}^{*}}$ are all $1$ so it is invertible. If however, $A$ has only one null eigenvalue then ${{I}_{n}}+{{A}^{*}}$ has $(n-1) $eigenvalues of 1 and the other one is the product of the nonzero eigenvalues of $A, + 1$. From this we deduce that $tr({{A}^{*}}) = \det(I+A^*) - 1$ and the equivalence is obvious. I hope it is clear what I said.

ionbursuc wrote: Given A, non-inverted matrices of order n with real elements, $n\ge 2$ and given ${{A}^{*}}$adjoin matrix A. Prove that $tr({{A}^{*}})\ne -1$ if and only if the matrix ${{I}_{n}}+{{A}^{*}}$ is invertible. Hello, can I know about the Math program in Romanian High School? Why do Romanian High School Students learn Higher Mathematics in grade 11?

Well, I can't speak for the Romanian Ministry of Education as to why the Romanian high school curriculum includes Higher Mathematics, but I can tell you exactly what is studied. In 11th grade we have linear algebra and differential calculus and in the 12th grade we have abstract algebra(basically groups and rings, stuff like that) and integral calculus.