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Let $A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)$. If $u_1$ and $u_2$ are column matrices such that $A u_1=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$ and $A u_2=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$, then $u_1+u_2$ is equal to
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$\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)$
$\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)$
$A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)$
Let $u_1=\left[\begin{array}{l}a \\ b \\ c\end{array}\right] ; u_2=\left[\begin{array}{l}d \\ e \\ f\end{array}\right]$
$\mathrm{Au}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] \quad \Rightarrow u_1=\left[\begin{array}{c}1 \\ -2 \\ 1\end{array}\right]$
$\mathrm{Au}_2=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right] \quad \Rightarrow \mathrm{u}_2=\left[\begin{array}{c}0 \\ 1 \\ -2\end{array}\right] \quad \Rightarrow \mathrm{u}_1+\mathrm{u}_2=\left[\begin{array}{c}1 \\ -1 \\ -1\end{array}\right]$
Let $u_1=\left[\begin{array}{l}a \\ b \\ c\end{array}\right] ; u_2=\left[\begin{array}{l}d \\ e \\ f\end{array}\right]$
$\mathrm{Au}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] \quad \Rightarrow u_1=\left[\begin{array}{c}1 \\ -2 \\ 1\end{array}\right]$
$\mathrm{Au}_2=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right] \quad \Rightarrow \mathrm{u}_2=\left[\begin{array}{c}0 \\ 1 \\ -2\end{array}\right] \quad \Rightarrow \mathrm{u}_1+\mathrm{u}_2=\left[\begin{array}{c}1 \\ -1 \\ -1\end{array}\right]$
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