Let $$A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}$$. If $$u_1$$ and $$u_2$$ are column matrices such that $$Au_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$ and $$Au_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$, then $$u_1 + u_2$$ is equal to

The given matrix is $$A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}$$. We must find column vectors $$u_1$$ and $$u_2$$ satisfying

$$A u_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \qquad A u_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.$$ Because $$A$$ is lower-triangular with all diagonal entries equal to $$1$$, each system can be solved quickly by back-substitution.

Case 1:

Let $$u_1 = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$. Then

$$ \begin{pmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}. $$

Row-wise equations:
$$a = 1,$$
$$2a + b = 0 \;\Longrightarrow\; b = -2,$$
$$3a + 2b + c = 0 \;\Longrightarrow\; 3 - 4 + c = 0 \;\Longrightarrow\; c = 1.$$

Thus $$u_1 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.$$

Case 2:

Let $$u_2 = \begin{pmatrix} d \\ e \\ f \end{pmatrix}$$. Then

$$ \begin{pmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} d \\ e \\ f \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}. $$

Row-wise equations:
$$d = 0,$$
$$2d + e = 1 \;\Longrightarrow\; e = 1,$$
$$3d + 2e + f = 0 \;\Longrightarrow\; 2 + f = 0 \;\Longrightarrow\; f = -2.$$

Thus $$u_2 = \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}.$$

Add the two vectors:

$$u_1 + u_2 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}.$$

Therefore, $$u_1 + u_2 = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}.$$
Option D which is: $$\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$$.