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

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. \(x + y + z = 10\) 3. \(x + y + z = 0\) ### Step 1: Write the Augmented Matrix The augmented matrix for the system can be constructed as follows: \[ \begin{bmatrix} -1 & 4 & 3 & | & -16 \\ 1 & 1 & 1 & | & 10 \\ 1 & 1 & 1 & | & 0 \end{bmatrix} \] ### Step 2: Row Reduction Next, we will perform row operations to reduce this matrix to row echelon form. 1. **Swap Row 1 and Row 2** to make the leading coefficient of the first row positive: \[ \begin{bmatrix} 1 & 1 & 1 & | & 10 \\ -1 & 4 & 3 & | & -16 \\ 1 & 1 & 1 & | & 0 \end{bmatrix} \] 2. **Add Row 1 to Row 2** and **subtract Row 1 from Row 3**: \[ \begin{bmatrix} 1 & 1 & 1 & | & 10 \\ 0 & 5 & 4 & | & -6 \\ 0 & 0 & 0 & | & -10 \end{bmatrix} \] ### Step 3: Analyze the Result The last row of the matrix is \(0 = -10\), which is a contradiction. This indicates that the system of equations is inconsistent. ### Conclusion Since we have reached a contradiction, we conclude that the system has no solution. Thus, the correct choice is: **C. This system has no solution.**