Let $A$ be a $n \times m$ matrix and $b$ be a $n \times 1$ vector (with real entries). Suppose the equation $Ax=b, x \in \Bbb R^m$ admits a unique solution. Then we can conclude that
$(\mathrm {a})$ $m \ge n$
$(\mathrm {b})$ $n \ge m$
$(\mathrm {c})$ $n=m$
$(\mathrm {d})$ $n>m$
How can I solve it? Please help me.
Thank you in advance.
In your question the number of equations is $n$ and the number of independent variables or unknowns is $m$. Now if $n<m$ then $\mathrm {rank}\ A \leq n$. So if you apply rank-nulity theorem you will get $\mathrm {rank}\ A + \mathrm {nulity}\ A = m \implies \mathrm {nulity}\ A = m - \mathrm {rank}\ A \geq m-n>0$. So the homogeneous undetermined system $Ax=0$ has infinitely many solutions. Now observe that if $x_0$ is a particular solution of the system $Ax=b$ and $h$ is a solution of $Ax=0$ then $x_0+h$ is also a solution of $Ax=b$. Since there are infinitely many choices for $h$ you have infinitely many solutions of the system $Ax=b$.
Now if you take the system $x+y+z=0;x+y+z=1$ then clearly the system has no solution. (Here the number of equations is fewer than the number of unknowns).
Conclusion $:$ If in a system of linear equations the number of equations is fewer than the number of unknowns (in your case if $n < m$) then either the system has no solution or has infinitely many solutions.
So for getting unique solution of the system we must have $n \geq m$. Hence $(b)$ is the only correct option.