A matrix with one row is a linear combination of two other rows is not invertible.

Question

I believe it's because once you start row reducing, one row will end up being all 0s, hence making the determinant of the matrix 0. If that is how you go about proving it, I am confused on how to do this abstractly, so it covers all the cases.

Answer 1

There are many options to approach this problem.

1) You can use the determinant as you describe. The answer does depend a bit on how you define the determinant (formula, alternating forms, volume). Since the row addition operation doesn't change the determinant, you know that the determinant of the matrix of interest is the same as the determinant of a matrix with a row of all zeros. Using cofactor expansion on the row of all zeros gives you a $0$ determinant.

2) You could use more of the definition of being invertible. Suppose that $R_k=aR_m+bR_n$ (following @David's notation). Then, if our matrix is $M$, then $M(\vec{e}_k)=M(a\vec{e}_m+b\vec{e}_n)$ where $\vec{e}_i$ represents the $i^{\text{th}}$ standard basis vector. Let's call $M(\vec{e}_k)=\vec{v}$. Now, suppose that $M$ had an inverse. Then $M^{-1}M=I$, so $$ \vec{e}_k=I\vec{e}_k=(M^{-1}M)\vec{e}_k=M^{-1}\vec{v} $$ and $$ (a\vec{e}_m+b\vec{e}_n)=I(a\vec{e}_m+b\vec{e}_n)=(M^{-1}M)(a\vec{e}_m+b\vec{e}_n)=M^{-1}\vec{v}. $$ Now, we have a contradiction because $M^{-1}\vec{v}$ can't equal two different things.

There are alternate approaches, but this should get you started.