I'd just like to verify that I have understood this correctly:
Let $B:=\{v_1,v_2,v_3\}$ with
$$v_1=\left(\begin{matrix}1\\1\\1 \end{matrix}\right),v_2=\left(\begin{matrix}1\\1\\0 \end{matrix}\right),v_3=\left(\begin{matrix}1\\0\\0 \end{matrix}\right).$$
a) Show this is a basis of $\mathbb{R}^3$.
b) Find a dual basis for $B$.
I think a) is clear to me. I test the vectors for linear independence, by bringing them in the form $$\left[\begin{array}{cccc|c}&1 &1 &1 &0\\&1 &1 &0 &0\\&1 &0 &0 &0 \end{array}\right]$$ and then solve this system of linear equations to show that the vectors are linearly independent since $x_1=x_2=x_3=0$.
Now to b). If I understand correctly, the dual basis, let's call it $D$, is biorthogonal to $B$ which can be expressed through $D^TB=I_3$, meaning that $D^T=B^{-1}$. So I simply calculate the inverse of B and then transpose the resulting matrix to get $D$? Is it this simple?
Try $$w_1=\left(\begin{matrix}0\\0\\1 \end{matrix}\right),w_2=\left(\begin{matrix}0\\1\\{-1} \end{matrix}\right),w_3=\left(\begin{matrix}1\\{-1}\\0 \end{matrix}\right).$$
You will have the biorthogonality and linear independence.