Consider the following linear equations
$1.x+0.y=2$
$0.x+1.y=3$
$0.x+0.y=3$
Obviously the above system has $0$ solutions.
But when you solve it in a matrix form
\begin{equation} \begin{bmatrix} 1 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}= \begin{bmatrix} 2\\ 3\\ 3 \end{bmatrix} \end{equation}
Multiply both sides of the equation with \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} which happens to be the left inverse of the co-efficient matrix, you get $x_1 = 2$ and $x_2 = 3$
Please give me an explanation to contradict this approach
Left-inverses prove injectivity, which isn't really what you are interested in here. You would need to find a right-inverse (which doesn't exist here) in order to prove the existence of a solution of a linear system for any right-hand-side.