# Is the system of linear equations always solvable if the matrix is of full rank

I know that for an $n×n$ matrix $A$, the system $Ax=b$ has a unique solution for all $b$ in $\mathbb{R}^{n}$. My question is : what will happen when $A$ is an $m\times n$ matrix?Does the system is solvable for every $b$ if rank of $A$ is $n$?

• Try the system $x=1, x=2$. On the other hand, if $m\leq n$ then yes, it must have solutions. – Hellen Oct 1 '17 at 12:42
• Didn't get ur answer.please explain – Mathlearner Oct 1 '17 at 12:45
• I hope you got that the system $x=1=2$ doesn't have solutions. For the second sentence, if $m\leq n$ then you can take $m$ variables corresponding to linearly dependent columns, give values to the remaining variables any way you want. Pass those terms to the other side, and the system is square. Therefore, as you know, it has a solution. – Hellen Oct 1 '17 at 12:48

No. If $A=[0, 1]^T$, then $Ax=[1, 0]^T$ has no solutions.
• Excepting the trivial solution $x=[0,0]^T$ – Paul Oct 1 '17 at 12:48
• @Paul $x\in \mathbb{R}$. – Koto Oct 1 '17 at 12:48
If the $m{\,\times\,}n\;$matrix $A\;$has rank $n$, the dimension of the column space is $n$, hence the dimension of $\text{im}(A)$ is $n$, so $\text{im}(A)$ is a proper subspace of $\mathbb{R}^m$, if $m >n$.
It follows that if $m > n$, there will always exist vectors $b \in \mathbb{R}^m$ such that the equation $Ax=b\;$is not solvable.
According to Rouché-Capelli Theorem, the linear system has solutions if and only if $A$ and the augmented matrix $A|b$ (obtained by adding $b$ to $A$ as a new column) have the same rank, say $r$. In that case the solutions are $\infty^{n-r}$, meaning that $r$ unknowns can be explicited in terms of the remaing $n-r$, which can be set to any value.
So, for example, if you take $A$ to be $m\times (m-1)$ and suppose (as you state in your question) that $A$ has rank $m-1$, than there are solutions iff the augmented matrix is singular (otherwise it would have rank $m$).