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.

To start, the augmented matrix for the given system of equations is: \[ \begin{pmatrix} -1 & 4 & 3 & | & -16 \\ 3 & 1 & 1 & | & 10 \\ 1 & 1 & 1 & | & 0 \end{pmatrix} \] Now, using the matrix method (specifically, row reduction), we can solve this matrix. The goal is to apply Gaussian elimination to bring the matrix to row-echelon form. 1. Start by swapping the first row with the third row to get a leading 1: \[ \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 3 & 1 & 1 & | & 10 \\ -1 & 4 & 3 & | & -16 \end{pmatrix} \] 2. Now, eliminate the x terms in the second and third rows. For the second row, we perform the operation \( R_2 - 3R_1 \): \[ \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & -2 & -2 & | & 10 \\ -1 & 4 & 3 & | & -16 \end{pmatrix} \] For the third row, we perform \( R_3 + R_1 \): \[ \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & -2 & -2 & | & 10 \\ 0 & 5 & 4 & | & -16 \end{pmatrix} \] 3. Next, we can simplify the second row by multiplying it by \(-\frac{1}{2}\): \[ \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 1 & 1 & | & -5 \\ 0 & 5 & 4 & | & -16 \end{pmatrix} \] 4. Now eliminate \( y \) from the third row by performing \( R_3 - 5R_2 \): \[ \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 1 & 1 & | & -5 \\ 0 & 0 & -1 & | & 9 \end{pmatrix} \] 5. Finally, multiply the third row by \(-1\): \[ \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 1 & 1 & | & -5 \\ 0 & 0 & 1 & | & -9 \end{pmatrix} \] Now back substitute to find the values: From Row 3: \( z = -9 \) From Row 2: \( y + z = -5 \) → \( y - 9 = -5 \) → \( y = 4 \) From Row 1: \( x + y + z = 0 \) → \( x + 4 - 9 = 0 \) → \( x = 5 \) Thus, the solution to the system is \( (5, 4, -9) \). The correct choice is A. This system has exactly one solution. The solution is (5, 4, -9).