Question given $A$ is $n×n$ real matrix & $b$ be an $n×1$ column vector. If the linear system $Ax=b$ has no solution then there are infinitely many vectors $c$ such that, $Ax=c$ has no solution.
My attempt let $k$ be any nonzero real number, consider the linear system $Ax=kb$. Let suppose that this linear system has a solution say $x_0$ then we have,
$$Ax_0=kb$$
$\implies \frac{1}{k}(Ax_0)=b$
$\implies A(\frac{1}{k}x_0)=b$
$\implies\frac{1}{k}x_0$ is a solution of the linear system $Ax=b$. Which contradicts the fact that, the linear system $Ax=b$ has no solution.
$\implies$ our assumption that, system $Ax=kb$ has a solution is wrong! Thus the linear system $Ax=kb$ has No solution! But $k≠0$ is arbitrary and hence there are infinitely many vectors $c=kb$ such that, the linear system $Ax=c$ has no solution. (Am i correct?)
Further, I didn't notice the use of non invertibility of $A$ in above? ( We know that,for $n×n$ matrix $A$ the system$Ax=b$ has no solution then the coefficient matrix $A$ is Not invertible ) Is we can prove the given statement by using the fact that $A$ is Not invertible?
Please help..
Yes @AkashPatalwanshi, your answer is correct. Since there's no $x$ such that $Ax = b$, there couldn't possibly be an $x'$ such that $Ax' = kb$ for any non-zero real number $k$ because that would imply that $Ax'\frac{1}{k} = b$, which would be a contradiction.
And we don't consider $k=0$ because $Ax= 0$ always has a trivial solution.
Edit (I just noticed that you were possibly looking for a proof using the non-invertibility of $A$)
Every possible $b$ here belongs to $\mathbb{R}^n$. However, since $A$ is non-invertible it strictly has less than $n$ linearly independent columns, or in other words, it's column space is at most $n-1$ dimensional, meaning there will always be vectors in $\mathbb{R}^n$ that aren't in the column space of $A$. This is because if every vector in $\mathbb{R}^n$ was in the column space of $A$, then the columns of $A$ would serve as a basis for $\mathbb{R}^n$, meaning there would have to be $n$ linearly independent columns in $A$ (because all bases of a vector space have the same cardinality) which is again a contradiction.
Hope it helps!