Let $A_{n \times n}$ be a matrix with entries from $\mathbb{R}$. And let $AX_{n \times 1} = B_{n \times 1}$ be a system of n linear equations, where $X$ is the vector of unknowns and $B \in \mathbb{R}^n$.
Let rank($A$) < $n$. Then show that:
$$B \in \text{Colspace}(A) \Leftrightarrow \text{adj}(A)\cdot B = 0 $$ where $\text{Colspace}(A)$ is the column space of $A$ and $\text{adj}(A)$ is the adjoint matrix i.e transpose of the cofactor matrix of $A$.
Showing the $\Rightarrow$ direction was easy: simply write $B$ as a linear combination of the columns of A to get the result. I am having difficulty proving the other way around.
I don't think that is true. If you take $A = \begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$, then $adj(A) = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$. For all $B\in\mathcal{M}_{3\times3}(\mathbb{R})$, $adj(A)\cdot B = 0$, but the rank of $A$ isn't $3$.
Let $A$ be a matrix. We call minors of $A$ the determinants of its square sub-matrices and minor of order k the determinant of a square submatrix of size $k$ obtained by removing $m - k$ rows and $n - k$ columns from the initial matrix, which we can denote det$A_{I, J}$, where $I$ (resp.$J$) is a part with $k$ elements of $\{1, ..., m $(resp, $n$)$\}$.
The Minors Theorem asserts that the rank of a matrix is equal to the largest order of a nonzero minor of that matrix, that is, to the integer r such that there is a nonzero minor of order r and that any minor of order strictly superior to r is null.
Note $adj(A)$ by $com(A)^{T}$. $Acom(A)^{T} = com(A)^{T}A =$ det$(A)\cdot I_{n}$, so if $A$ isn't inversible, we have $Im(A)\subset Ker(com(A)^{T})$ and $Im(com(A)^{T})\subset Ker(A)$.
If the rank of matrix $A$ is strictly inferior to $n-1$, $com(A)^{T}$ equals to $0$. If the rank of matrix $A$ equals to $n-1$, since $com(A)^{T}$ is nonzero, dim $Ker(com(A)^{T}) < n$, then we have dim $Im(com(A)^{T}) = 1$.
By dimension, we can conclude that if $rank(A) = n-1$, $Im(A) = Ker(com(A)^{T})$, i.e $B \in \text{Colspace}(A) \Leftrightarrow \text{adj}(A)\cdot B = 0$