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The set of vectors are:

S={ (1,6,4) , (2,4,1) , (-1,2,5) }

for it to be a base in $\mathbb{R}^3$ it must be linearly independent and span $\mathbb{R}^3$.

I first tried to solve it by taking its determinant and found that it's not zero, and therefore the set is linearly independent. Now how would I show that is spans? since none of them are scalar multiples of each other?

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  • $\begingroup$ Just form the matrix with rows equal to each vector. If the determinant of this matrix is $0$, then they are linearly dependent and do not span $\mathbb{R}^3$. Else, they are linearly independent and do span $\mathbb{R}^3$. $\endgroup$ – Guest Apr 1 '14 at 2:29
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The Fundamental Theorem of Invertible Matrices says, among several things, that the statement "$A$ is an invertible $n\times n$ matrix" is logically equivalent to the statement "The rows (and columns) of $A$ span $\mathbb{R}^n$, are linearly independent, and form a basis for $\mathbb{R}^n$."

So representing your vectors as a 3x3 matrix (and showing it is invertible with a nonzero determinant) you can use the Fundamental Theorem of Invertible Matrices and you're done!

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Make a matrix $A$ by copying the given vectors as it columns.

For any given column vecor $b\in\mathbf{R}^3$, the linear $A(x\ y\ z)^T=b$ is solvable uniquely as you have verified the determinant is non-zero. (The solution vector is simply $A^{-1}b$.) The solution values of $x,y,z$ provide the coefficients forming the linear combination of elements of $S$ giving (the arbitrary) $b$. So they span.

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