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Let $A$ be a $n \times n$ square matrix whose all columns are independent. Is $Ax = b$ always solvable?

Actually, I know that $Ax= b$ is solvable if $b$ is in the column space of $A$. However, I am not sure if it is solvable for all values of $b$.

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Few Inferences first :

  • If the Rank of the Augmented Matrix $[A:b]$ $=$ Rank $[A]$ =$N$(Order of the matrix A) then it is solvable and can be solved by reducing by using Gaussian Elimination Method, and it has Unique Solution.
  • If the  Rank of the Augmented Matrix $[A:b]$  $=$ Rank $[A]$$ < N$ then it is solvable and has Infinitely many solutions.
  • If the  Rank of the Augmented Matrix $[A:b]$  $\neq$ Rank $[A]$ then it is not solvable.

Now as per your question, if A is a vector whose all columns are linearly independent then in that case it has a rank = $n$ and whatever may be $b$ the augmented matrix $[A:b]$ will always have a rank = $n$ hence it is always solvable and will have unique solution.

For more details you can refer to the lectures by Gilbert Strang or his book on linear algebra.

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