Is this set a basis for $R^3$. How to verify? Say I have $S = {(1,0,-1), (2,1,1), (-3,0,2)}$. Is my method correct for determining if it's a basis? How do I check if I'm right?
We need to determine if $S $ spans $R^3$ and if it's linearly independent.
First, check if $S$ spans $R^3$:
Let $u_1, u_2, u_3$ be a random vector.
$$(u_1, u_2, u_3) = c_1(1,0,-1) + c_2(2,1,1) + c_3(-3,0,-2)$$
$$ = (c_1 + 2c_2 - 3c_3, c_2, -c_1 + c_2 + 2c_3$$
the corresponding equations:
$$c_1 + 2c_2 - 3c_3 = u_1$$
$$c_2 = u_2$$
$$-c_1 + c_2 + 2c_3 = u_3$$
and the resulting matrix:
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ -1 & 1 & 2 \end{bmatrix}$$
Determine determinant using expansion by cofactors in the second row:
determinant = $0$ * (cofactor of 2nd row, 1st column) + 1 * (cofactor of 2nd row, 2nd column) + $0$ * (cofactor of 2nd row, third column) = 5
So the fact that the determinant is non-zero means that there is a unique solution for this system of equations for any random vector in $R^3$
Is it linearly dependent?
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ -1 & 1 & 2 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ 0 & 3 & -1 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$
Because the last row is $[0,0,1]$ this implies that the only solution is trivial so it is linearly dependent.
So $S $ is a basis for $R^3$. Is this logic right?

Take a look at the zero-components.  The second vector can’t be a linear combination of the other vectors.  Since these other vectors  aren’t parallel, the three vectors are linearly independent.
