linear equation $x^TA =b$ How do you solve the following linear equation: $x^TA = b$?  
Does a solution exist, and is it unique? I know how to solve $Ax = b$, but what about $x^TA = b$?
Thanks
 A: Take transpose.
$$A^Tx=b^T$$
Can you solve the problem now?
A: The following discussion pertains to the case $A$ invertible:
If you start with the equation
$x^TA = b, \tag{1}$
you can simply right multiply by $A^{-1}$:
$x^T =  x^T I = x^T(AA^{-1}) = (x^TA)A^{-1} = bA^{-1}; \tag{2}$
this works of course because $b$ is a row-vector.
We could of course also first take the transpose of (1):
$A^Tx = (x^TA)^T = b^T, \tag{3}$
and then left multiply by $(A^T)^{-1}$:
$x = Ix = ((A^T)^{-1}A^T)x = (A^T)^{-1}(A^Tx) = (A^T)^{-1}b^T. \tag{4}$
We can compare/validate the solution(s) by taking the transpose of (2):
$x = (x^T)^T = (bA^{-1})^T = (A^{-1})^Tb^T; \tag{5}$
(4) and (5) will evidently agree if
$(A^T)^{-1} = (A^{-1})^T; \tag{6}$
but since
$A^{-1}A = I, \tag{7}$
we have
$A^T(A^{-1})^T = I^T = I, \tag{8}$
thus
$(A^{-1})^T = (A^T)^{-1}; \tag{9}$
hence (4) and (5) agree.
So you need to know how to compute matrix inverses for $A$ and $A^T$, a standard problem.
One can of course apply other methods to (1), (3), such as Gaussian elimination etc.  But here we see why the results will agree.
