consider a linear system as follows
$a_{1,1}\ x_1+a_{1,2}\ x_2+…+a_{1,n}\ x_n = b_1$
$a_{2,1}\ x_1+a_{2,2}\ x_2+…+a_{2,n}\ x_n = b_2$
…
$a_{m,1}\ x_1+a_{m,2}\ x_2+…+a_{m,n}\ x_n = b_m$
in array form,
$A_{m \times n} x_{n \times 1} = b_{m \times 1}$
can someone give me the conditions for this system to be
1) consistent, i.e. atleast one solution exists
2) informationally complete, i.e. there is exactly one unique solution
Also is it possible for $A$ to be invertible with inverse $A_{n \times m}^{-1}$
but $x_{n \times 1} \neq A_{n \times m}^{-1}\ b_{m \times 1}$.
Please i am stuck in a problem which requires this result. I donot need proof, just the general condition for consistency and uniqueness of solution in a Linear system.
Consider the following example,
$A:= [[3/2,1/2],[3/2,3/2],[3/2,3]]$
$b:= [2,0,1]^T$
now $A^{-1}$ exists and is $[[12/19,16/57,-14/57],[-7/19,-1/19,8/19]]$
but $A^{-1}b = [58/57,-6/19]^T$
what is $A^{-1} b$ here? since the system is inconsistent with no solution.
Best Answer
It is not possible for $A$ to be invertible with inverse $A_{n \times m}^{-1}$ but $x_{n \times 1} \neq A_{n \times m}^{-1}\ b_{m \times 1}$.
1) The system is consistent when $A$ and $A$ extended with $b_m$ as another column has the same rank.
If the system is consistent then
if $n > m$ it is not informationally complete,
if $n = m$,
2) The linear system is informationally complete when $A$ is invertible. When it isn't, the system either has infinite solutions or none.
if $n < m$ you can get rid of the redundant equations by Gauss Elimination and get back to the other two cases.