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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.

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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.

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Hint: Your equation $Ax=b$ is a set of linear simultaneous equations. How many variables are there? How many equations are there?

Looking at the choices, you know $n=m$ is a possibility. That rules out d.

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Let $A = <a_{i,j}>$ and $x = (x_1,x_2, .... , x_n)$ and $b = (b_1, b_2, .... b_m)$

So $A\times x = b$ means we have $m$ linear equations.

$a_{1,1}x_1 + a_{2,1}x_2 + .... +a_{n,1}x_n = b_1$

$a_{1,2}x_1 + a_{2,2}x_2 + .... +a_{n,2}x_n = b_2$

.....

$a_{1,m}x_1 + a_{2,m}x_2 + .... +a_{n,m}x_n = b_m$

These $m$ equations with $n$ unknowns have a unique solution.

Is that possible if $m < n$? Is that possible if $m= n$? Is that possible if $m > n$?

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  • $\begingroup$ Nothing can be said. $\endgroup$ – Dbchatto67 May 19 '18 at 15:57
  • $\begingroup$ No. Plenty can be said. think about it. If you have $5x + 3y + 2z = 7$ and $9x + 2y + 4z = 9$ can there be unique solutions? (Try and solve it). That is exactly what these would be if $n=3;m =2$ and $A$ was the 3x2 matrix with (5,3,2) in top row and (9,2,4) in bottom row and $b = (7,9)$ and we where trying to solve for $x = (x,y,z)$ and hoping there is only one solution. $\endgroup$ – fleablood May 19 '18 at 16:18
  • $\begingroup$ Consider the Linear Transformation. $T:\mathbb R^m \to \mathbb R^n$. $T(x)=Ax=b$. In the question given that $x$ is unique. So, $T$ should be bijective. right?. When $m\ge n$, we can conclude that $T$ is bijective. For the uniqueness of $x$. $n\ge m$. right? $\endgroup$ – user464147 Nov 17 '19 at 16:25
  • $\begingroup$ No.... a unique solution does not imply bijectivity and $m\ge n$ neither implies a unique solution (the vectors need not be independant) nor bijective. But $m < n$ guarantees not bijective and no unique solutions. $\endgroup$ – fleablood Nov 17 '19 at 16:34

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