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I have to show that the real matrix

$$ A = \begin{pmatrix} 2 & -2 & 1 \\ 1 & -2 & 1 \\ -2 & 3 & -1 \\ \end{pmatrix} $$ Is invertible and use this to conclude that the elements:

$$ \begin{pmatrix} 2 \\ 1 \\ -2 \\ \end{pmatrix}, \begin{pmatrix} -2 \\ -2 \\ 3 \\ \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ -1 \\ \end{pmatrix} $$ span the vector space $\mathbb {R}^3$.

In order to prove that the matrix is invertible I know if I can find an inverse matrix B, such that $A*B=I_n $ then the matrix is invertible - I have not learned about the determinant method yet, so I cannot use that method. So when I know that A is invertible how does that imply that the elements (the columns in A) span the vector space?

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If your matrix is invertible hence it has a full rank therefore its columns are linearly independent and form a basis in $\mathbb{R}^3$

Update: If you are not familiar with such concept as basis we can solve this problem in other way: since your matrix is invertible hence $\forall \ b \in \mathbb{R}^3 \ \exists \ x : Ax = b$ and this $x$ is equal to $A^{-1}b$. Since $Ax$ means a linear combination of the columns of $A$ with coefficients $x_1, x_2, x_3$ where $x = (x_1 \ x_2 \ x_3)^T$ then every vector from $\mathbb{R}^3$ can be represented as the linear combination of the columns of your matrix therefore its columns span the $\mathbb{R}^3$

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  • $\begingroup$ That is a method that I am more familiar with using at this point, thank you every much for you help! $\endgroup$ – Simbörg Feb 18 '18 at 14:54
  • $\begingroup$ @Simbörg You're welcome! $\endgroup$ – D F Feb 18 '18 at 14:56

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