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Take $B$ to be any $n \times n$ square matrix. If $A$ is an invertible $n \times n$ matrix, then does the set of $AB$ span all $n \times n$ matrices?

Edit: I believe it is true because for a given matrix $C$ you are trying to construct from the product $AB$, you know the entries of $A$ and you know the entries of $C$ so you can construct a system of equations to find the entries of $B$.

Could someone clarify if this reasoning is okay?

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Let $C$ be a matrix. Set $B=A^{-1}C$. Then $AB=C$.

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  • $\begingroup$ awesome, just making sure! is my justification right as well? $\endgroup$
    – atul ganju
    Commented May 15, 2020 at 22:54
  • $\begingroup$ Yes. If I understand correctly, you are solving the linear system $AX=C$, and solving this system gives $B$. This is one way to interpret $B=A^{-1}C$. $\endgroup$
    – Isaac Ren
    Commented May 15, 2020 at 22:56

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