How to construct a matrix given the null basis of A? Construct a $4\times4$ matrix $A$ such that $\{(1,2,3,4),(1,1,2,2)\}$ is a basis of $N(A)$.
So I know that $A$ will have two pivot columns and two free columns, but beyond this I'm not sure how to approach/solve.

Let me show you how it would go. Call $v_1=(1,2,3,4)$ and $v_2=(1,1,2,2)$. 
Let $v=(x_1,x_2,x_3,x_4)$. We want $v\cdot v_1=0$ and $v\cdot v_2=0$. This gives us the system of equations $$\begin{align}x_1+2x_2+3x_3+4x_4&=0\\x_1+x_2+2x_3+2x_4&=0\end{align}$$
Subtracting the first equation from the second we get: $$\begin{align}x_1+2x_2+3x_3+4x_4&=0\\\phantom{x_1}-x_2-x_3-2x_4&=0\end{align}$$
Now the system is in echelon form. In order to get two linearly independent solutions we can put first $x_3=1$, $x_4=0$ and solve for $x_1,x_2$, and then put $x_3=0$, $x_4=1$ and solve for $x_1,x_2$.
Doing that gives us two vectors $$\begin{align}u_1&=(-1,-1,1,0)\\u_2&=(0,-2,0,1)\end{align}$$
Therefore, one matrix satisfying the conditions of the problem would be $$\begin{pmatrix}-1,&-1,&1,&0\\0,&-2,&0,&1\\-1,&-1,&1,&0\\0,&-2,&0,&1\end{pmatrix}$$
obtained by putting $u_1,u_2$ and again $u_1,u_2$ as rows.
