I'm having real trouble understanding this question. I am supposed to figure out if the following statement is true or false, and have been thinking for hours but I cannot seem to think of an answer for this.. Much thanks to anyone who'd be able to explain the following question..!
If $A$ is an $n\times n$ matrix with $2$ linearly dependent rows, then the adjoint of $A$ has at least $n − 2$ zero columns.
The adjoint matrix $\operatorname{adj}A$ has entries $$ b_{ij}=(-1)^{i+j}\det A_{ji} $$ where $A_{ji}$ is obtained by removing the $j$-th row and $i$-th column from $A$.
It is not restrictive to assume that the linearly dependent rows of $A$ are the two at the bottom.
Then any determinant that is obtained by removing the $j$-th row where $1\le j\le n-2$ is zero, because the matrix still has two linearly dependent rows. Thus the first $n-2$ columns of $\operatorname{adj}A$ are zero.