In an augmented matrix representing a system of equations, why is it a contradiction when the LHS isn't zero and RHS is zero but not when flipped?

Question

In an augmented matrix representing a system of equations, say a $1\times 3$ matrix: $(a,b \mid c)$, why is it a contradiction when $a=b=0, c\neq 0$ but not when $a,b\neq 0, c=0$ ?

Answer 1

I guess you are talking about augmented matrices for systems of linear equations. Such a system is described by a matrix equation $Ax=c$, where $A$ is an $m\times n$ matrix, $b$ a column vector of $m$ rows and $x$ (column vector of $n$ rows) are the solutions you are looking for. You write this system just as $(A\mid c)$. Now when $A=0$ and $c\neq 0$, we are dealing with $0x=c$, which can't have any solutions, since $0x$ (where $0$ is the $m\times n$ zero matrix) is always $0$ (the column vector of $m$ zeros), so can't be equal to $c$ being non-zero.

On the other hand, when $A\neq 0$ but $c=0$ you can indeed have solutions. For example $$ \pmatrix{1&0\\0&0} \pmatrix{0\\1} = \pmatrix{0\\0}. $$