# $\forall \vec{b} \in \mathbb{R}^n, B \vec{x} = \vec{b}$ is consistent is equivalent to…

Suppose $A: n \times n$ and $B: n \times m$ and that $A$ is
invertible.

Prove that $\forall \vec{b} \in \mathbb{R}^n, B \vec{x} = \vec{b}$ is
consistent is equivalent to $\forall \vec{b} \in \mathbb{R}^n, (AB) \vec{x} = \vec{b}$ is consistent.

I’m assuming this boils down to proving that $B$ and $AB$ are equivalent expressions, but I can’t see why this would be the case. Why does $A$ being invertible make the two statements equivalent (that is, why is $A$ invertible required)?

#### Solutions Collecting From Web of "$\forall \vec{b} \in \mathbb{R}^n, B \vec{x} = \vec{b}$ is consistent is equivalent to…"

Prove that $b\in$ column space of $B$ if and only if $A^{-1}b\in$ column space of $B$, which is pretty obvious. Because if $b_1,\cdots,b_m$ are the columns of $B$, then $b$ can written as a linear combination, $c_1b_1+\cdots+c_mb_m$ implies $A^{-1}B$ can be written as $c_1A^{-1}b_1+\cdots+c_mA^{-1}b_m$ and vice versa. Note that as $A$ is invertible $A^{-1}b_1,\cdots,A^{-1}b_m$ is a basis for the column space of $B$.