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Suppose a matrix A is such that (i) the matrix-vector equation Ax = (1,1,1)has a unique solution;

I am trying to find the dimension of the row space of A, and apparently (i) tells me that the null space is 0, therefore I can use the rank nullity theorem. Why does (i) tell me that the null space is 0 though?

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Because if $u\in\operatorname{Null}(A)\setminus\{0\}$ and $v$ is such that $A.v=(1,1,1)$, then$$A.(v+u)=A.v+A.u=(1,1,1)+(0,0,0)=(1,1,1),$$but $u+v\neq v$.

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