Let \(A=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\). If \(u_1\) and \(u_2\) are column matrices such that \(Au_1 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\) and \(Au_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\), then \(u_1 - u_2\) is
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If \(Au_1\) and \(Au_2\) are known, then \(A(u_1-u_2)=Au_1-Au_2\) because matrix multiplication spreads over subtraction. So we first find the right hand side, then solve a simple system.
Step 1: Combine the two given facts. We know \(Au_1\) and \(Au_2\). Since A acts linearly, \(A(u_1-u_2)=Au_1-Au_2\).
Step 2: Subtract the right hand sides. \[Au_1-Au_2=\begin{bmatrix}2\\1\\0\end{bmatrix}-\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\\-1\end{bmatrix}.\] Step 3: Let \(v=u_1-u_2\) and write the equation. So \(Av=\begin{bmatrix}1\\1\\-1\end{bmatrix}\), where \(A=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}\).
Step 4: Solve from the bottom row up. The last row gives \(v_3=-1\).
Step 5: Use the middle row. \(v_2+v_3=1\Rightarrow v_2-1=1\Rightarrow v_2=2\).
Step 6: Use the top row. \(v_1+v_2=1\Rightarrow v_1+2=1\Rightarrow v_1=-1\). So \(u_1-u_2=\begin{bmatrix}-1\\2\\-1\end{bmatrix}\), which is option 1. \[ \boxed{u_1-u_2=\begin{bmatrix}-1\\2\\-1\end{bmatrix}} \]