Let `A=((1,0,0),(2,1,0),(3,2,1))`. If `u_(1)` and `u_(2)` are column matrices such that `Au_(1)=((1),(0),(0))` and `Au_(2)=((0),(1),(0))`, then `u_(1)+u_(2)` is equal to :

Options

A. `((-1),(1),(0))`
B. `((-1),(1),(-1))`
C. `((-1),(-1),(0))`
D. `((1),(-1),(-1))`

Explanation

To solve the problem, we need to find the sum of the column matrices \( u_1 \) and \( u_2 \) given that:

1. \( A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix} \)
2. \( A u_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \)
3. \( A u_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \)

### Step 1: Find \( u_1 \)

We start by solving for \( u_1 \) using the equation \( A u_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \).

Let \( u_1 = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \).

Then, we have:

\[ A u_1 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]

This leads to the following system of equations:

1. \( 1 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 = 1 \) → \( x_1 = 1 \)
2. \( 2 \cdot x_1 + 1 \cdot x_2 + 0 \cdot x_3 = 0 \) → \( 2 \cdot 1 + x_2 = 0 \) → \( x_2 = -2 \)
3. \( 3 \cdot x_1 + 2 \cdot x_2 + 1 \cdot x_3 = 0 \) → \( 3 \cdot 1 + 2 \cdot (-2) + x_3 = 0 \) → \( 3 - 4 + x_3 = 0 \) → \( x_3 = 1 \)

Thus, we find:

\[ u_1 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \]

### Step 2: Find \( u_2 \)

Next, we solve for \( u_2 \) using the equation \( A u_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \).

Let \( u_2 = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \).

Then, we have:

\[ A u_2 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \]

This leads to the following system of equations:

1. \( 1 \cdot y_1 + 0 \cdot y_2 + 0 \cdot y_3 = 0 \) → \( y_1 = 0 \)
2. \( 2 \cdot y_1 + 1 \cdot y_2 + 0 \cdot y_3 = 1 \) → \( 0 + y_2 = 1 \) → \( y_2 = 1 \)
3. \( 3 \cdot y_1 + 2 \cdot y_2 + 1 \cdot y_3 = 0 \) → \( 0 + 2 + y_3 = 0 \) → \( y_3 = -2 \)

Thus, we find:

\[ u_2 = \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix} \]

### Step 3: Calculate \( u_1 + u_2 \)

Now we can find \( u_1 + u_2 \):

\[ u_1 + u_2 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 + 0 \\ -2 + 1 \\ 1 - 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \]

### Final Answer

Thus, \( u_1 + u_2 = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \).

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