Suppose square matrix $A$ whose size are $n$ x $n$ having a determinant equals to $0$. Is it possible to convert it into invertible matrix by setting maximum three columns into any specific values?
If so, What supposed to be the value of those columns ?
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In general, as previously described, the main concept is rank, which is the dimension of the space spanned by A's columns.
If $rank(A) >n-3$ then it has at most three columns linearly dependent on the rest, if we replace them with independent one, we get an invertible matrix(because all whose columns are linearly independent).
else, it cannot be done, as we have four or more such columns that can not all be replaced.
If I understood well, then no. Imagine following matrix \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} Then you cant convert into invertible matrix by setting maximum three columns. If rank(A)=n-3, then maybe.