So I'm asked to proof for any matrix that complies with one of the following rules, the determinant of that matrix is $0$.
- Two columns of a matrix A are the same
- A row of a matrix A is a scalar multiple of another row of matrix A
- The sum of two rows of a matrix A equals a third row of the matrix A
I've been able to prove these rules for both $2\times2$ and a $3 \times 3$ matrices (only $3 \times 3$ in case of the last one), but I'm now questioning how I can extend this to be proven for any $n \times n$ matrix. Is the best option I have here induction or is there an other way to elegantly prove this.
I will assume that you have already proven that if the determinant of a matrix is non-zero, then that matrix has an inverse. Recall that if $M$ is invertible, then $Mx = 0$ implies that $x = 0$. With that, note the following:
Two columns of $A$ are the same if and only if there is a non-zero column-vector $x$ for which $Ax = 0$, and $x$ the entries of $x$ are all zero except for one entry equal to $1$ and another equal to $-1$.
One row is a scalar multiple of the other if and only if there is a non-zero column-vector $x$ for which $A^Tx = 0$ (where $A^T$ denotes the transpose of $A$), and $x$ has only $2$ non-zero entries.
The sum of two rows is equal to a third if and only if there is a non-zero column vector $x$ for which $A^Tx = 0$, and $x$ has exactly $3$ non-zero entries, with one of those non-zero entries equal to $1$.
We could prove this as a consequence of the effects of elementary row-operations and column-operations on the determinant of the matrix. In particular, note that