# Determinant of matrix relations

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.

• What definition do you have for the determinant of a matrix? Jun 3, 2020 at 14:51
• Have you already proven that if the determinant of a matrix is non-zero, then that matrix has an inverse? Jun 3, 2020 at 14:52
• @Gribouillis, the definition of a matrix here is: $$\begin{pmatrix} a_{11} & a_{12} & ... a_{1n} \\ a_{21} & a_{22} & ... a_{2n} \\ ... & ... & ... ... \\ a_{m1} & a_{m2} & ... a_{mn} \\\end{pmatrix}$$ Jun 3, 2020 at 14:57
• @Omnomnomnom yes I have proven that already Jun 3, 2020 at 14:59

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

• Switching two identical columns must change the sign of the determinant, yet the determinant must remain the same,
• Subtracting a scalar multiple of one row from the other (to produce a zero-row) does not alter the determinant,
• Subtracting one row from the other (to produce a repeated row) does not alter the determinant.