Find the null space of the following matrix. \[ A = \begin{bmatrix} 1 & 2 & -1 \ 2 & 4 & -2 \ -1 & -2 & 1 \end{bmatrix} \]

Step 1 ：Step 1: Set the matrix equal to the zero vector and solve for the free variables. \[ \begin{bmatrix} 1 & 2 & -1 \ 2 & 4 & -2 \ -1 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \]

Step 2 ：Step 2: This gives us the system of equations \(x + 2y - z = 0\), \(2x + 4y - 2z = 0\), and \(-x - 2y + z = 0\). It can be observed that the second equation is just twice the first equation, and the third equation is just the negative of the first equation. Hence, we can reduce the system of equations to just \(x + 2y - z = 0\).

Step 3 ：Step 3: Rearranging the equation we get \(x = z - 2y\). We can set \(y = t\) and \(z = s\), where \(t\) and \(s\) are any real numbers, to get the general solution \(x = s - 2t\), \(y = t\), \(z = s\).