Write the augmented matrix of the system and use the matrix method to solve the system. $$ \begin{array}{l}-x+4 y+3 z=-16 \ 3 x+y+z=10 \ x+y+z=0\end{array} $$ Select the correct choice and, if necessary, fill in the answer boxes to complete your choice. A. This system has exactly one solution. The solution is ( (Type integers or decimals.) B. This system has infinitely many solutions of the form ( $$ \square, \square, z) $$, where $$ z $$ is any real number. (Type integers or decimals.) C. This system has no solution.

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To solve the given system of equations using the matrix method, we first need to write the augmented matrix for the system:

The system of equations is:
1. $$-x + 4y + 3z = -16$$
2. $$3x + y + z = 10$$
3. $$x + y + z = 0$$

The augmented matrix for this system can be represented as follows:

$$
\begin{bmatrix}
-1 & 4 & 3 & | & -16 \
3 & 1 & 1 & | & 10 \
1 & 1 & 1 & | & 0
\end{bmatrix}
$$

Next, we will use row operations to reduce this augmented matrix to row echelon form or reduced row echelon form.

### Step 1: Row Operations

1. **Swap Row 1 and Row 3** to make the leading coefficient of the first row positive:
   $$
   \begin{bmatrix}
   1 & 1 & 1 & | & 0 \
   3 & 1 & 1 & | & 10 \
   -1 & 4 & 3 & | & -16
   \end{bmatrix}
   $$

2. **Eliminate the first column below Row 1**:
   - Row 2: $$R_2 - 3R_1$$
   - Row 3: $$R_3 + R_1$$

Performing these operations gives:
   $$
   \begin{bmatrix}
   1 & 1 & 1 & | & 0 \
   0 & -2 & -2 & | & 10 \
   0 & 5 & 4 & | & -16
   \end{bmatrix}
   $$

3. **Simplify Row 2** by dividing by -2:
   $$
   \begin{bmatrix}
   1 & 1 & 1 & | & 0 \
   0 & 1 & 1 & | & -5 \
   0 & 5 & 4 & | & -16
   \end{bmatrix}
   $$

4. **Eliminate the second column below Row 2**:
   - Row 3: $$R_3 - 5R_2$$

Performing this operation gives:
   $$
   \begin{bmatrix}
   1 & 1 & 1 & | & 0 \
   0 & 1 & 1 & | & -5 \
   0 & 0 & -1 & | & 9
   \end{bmatrix}
   $$

5. **Simplify Row 3** by multiplying by -1:
   $$
   \begin{bmatrix}
   1 & 1 & 1 & | & 0 \
   0 & 1 & 1 & | & -5 \
   0 & 0 & 1 & | & -9
   \end{bmatrix}
   $$

### Step 2: Back Substitution

Now we can back substitute to find the values of $$x$$, $$y$$, and $$z$$:

From Row 3:
$$
z = -9
$$

From Row 2:
$$
y + z = -5 \implies y - 9 = -5 \implies y = 4
$$

From Row 1:
$$
x + y + z = 0 \implies x + 4 - 9 = 0 \implies x = 5
$$

### Final Solution

The solution to the system is:
$$
(x, y, z) = (5, 4, -9)
$$

### Conclusion

Since we found a unique solution, the correct choice is:
**A. This system has exactly one solution. The solution is (5, 4, -9).**
The system has exactly one solution: $$ x = 5 $$, $$ y = 4 $$, $$ z = -9 $$.

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