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Isn't this wrong? This is inconsistent and therefore S does not generate R^3.

Right?

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1 Answer

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It seems a perfect solution.

The row reductions correspond to elementary basis transformations on $\Bbb R^3$: starting out from the standard basis $e_1=\pmatrix{1\\0\\0}$, $e_2=\pmatrix{0\\1\\0}$, $e_3=\pmatrix{0\\0\\1}$, we do the following similar (inverse) steps to the basis, and coordinate the same four vectors in the new basis:

• $R_2\,\to\, R_2+3\,R_1$ corresponds to $e_1' :=e_1-3e_2$ (so that, indeed the first vector becomes $-e_1'$).
• $R_3\,\to\,R_3-R_2$ corresponds to $e_2':=e_2+e_3$.

Since these always remain basis, the vectors $-e_1',\,e_2',2e_1'+5e_2'-4e_3$ also form a basis, in particular they span all vectors of $\Bbb R^3$.

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