Use Gaussian elimination to find the complete solution to the system of equa \[ \begin{array}{c}x+3 y-5 z-w=11 \\ 4 x+y+3 z+3 w=1 \\ -3 x-y+z-2 w=-7 \\ x-y-3 z-3 w=-1\end{array} \] Select the correct choice below and, if necessary, fill in the answer boxes to A. There is one solution. The solution set is \( \{\square \). (Simplify your answers.) B. There are infinitely many solutions. The solution set is \( \{\square \). (Simplify your answers. Type expressions using \( w \) as the variable. C. There is no solution. The solution set is \( \varnothing \).

To solve the system of equations using Gaussian elimination, we first write the augmented matrix corresponding to the system: \[ \begin{bmatrix} 1 & 3 & -5 & -1 & | & 11 \\ 4 & 1 & 3 & 3 & | & 1 \\ -3 & -1 & 1 & -2 & | & -7 \\ 1 & -1 & -3 & -3 & | & -1 \end{bmatrix} \] Next, we will perform row operations to bring this matrix to row echelon form. 1. **Row 2**: Replace \( R_2 \) with \( R_2 - 4R_1 \): \[ R_2 = [4, 1, 3, 3 | 1] - 4[1, 3, -5, -1 | 11] = [0, -11, 23, 7 | -43] \] 2. **Row 3**: Replace \( R_3 \) with \( R_3 + 3R_1 \): \[ R_3 = [-3, -1, 1, -2 | -7] + 3[1, 3, -5, -1 | 11] = [0, 8, -14, -5 | 26] \] 3. **Row 4**: Replace \( R_4 \) with \( R_4 - R_1 \): \[ R_4 = [1, -1, -3, -3 | -1] - [1, 3, -5, -1 | 11] = [0, -4, 2, -2 | -12] \] Now our matrix looks like this: \[ \begin{bmatrix} 1 & 3 & -5 & -1 & | & 11 \\ 0 & -11 & 23 & 7 & | & -43 \\ 0 & 8 & -14 & -5 & | & 26 \\ 0 & -4 & 2 & -2 & | & -12 \end{bmatrix} \] Next, we will continue to simplify the matrix. 4. **Row 2**: Divide \( R_2 \) by -11: \[ R_2 = [0, -11, 23, 7 | -43] \rightarrow [0, 1, -\frac{23}{11}, -\frac{7}{11} | \frac{43}{11}] \] 5. **Row 3**: Replace \( R_3 \) with \( R_3 - 8R_2 \): \[ R_3 = [0, 8, -14, -5 | 26] - 8[0, 1, -\frac{23}{11}, -\frac{7}{11} | \frac{43}{11}] = [0, 0, \frac{2}{11}, \frac{1}{11} | -\frac{2}{11}] \] 6. **Row 4**: Replace \( R_4 \) with \( R_4 + 4R_2 \): \[ R_4 = [0, -4, 2, -2 | -12] + 4[0, 1, -\frac{23}{11}, -\frac{7}{11} | \frac{43}{11}] = [0, 0, -\frac{6}{11}, -\frac{1}{11} | -\frac{1}{11}] \] Now our matrix looks like this: \[ \begin{bmatrix} 1 & 3 & -5 & -1 & | & 11 \\ 0 & 1 & -\frac{23}{11} & -\frac{7}{11} & | & \frac{43}{11} \\ 0 & 0 & \frac{2}{11} & \frac{1}{11} & | & -\frac{2}{11} \\ 0 & 0 & -\frac{6}{11} & -\frac{1}{11} & | & -\frac{1}{11} \end{bmatrix} \] 7. **Row 3**: Multiply \( R_3 \) by \( \frac{11}{2} \): \[ R_3 = [0, 0, 1, \frac{1}{2} | -1] \] 8. **Row 4**: Replace \( R_4 \) with \( R_4 + 3R_3 \): \[ R_4 = [0, 0, 0, \frac{1}{2} | -\frac{5}{2}] \] Now our matrix looks like this: \[ \begin{bmatrix} 1 & 3 & -5 & -1 & | & 11 \\ 0 & 1 & -\frac{23}{11} & -\frac{7}{11} & | & \frac{43}{11} \\ 0 & 0 & 1 & \frac{1}{2} & | & -1 \\ 0 & 0 & 0 & 1 & | & -5 \end{bmatrix} \] Now we can back substitute to find the solutions. From the last row, we have: \[ w = -5 \] Substituting \( w = -5 \) into the third row: \[ z + \frac{1}{2}(-5) = -1 \implies z - \frac{5}{2} = -1 \implies z = \frac{3}{2} \] Substituting \( z = \frac{3}{2} \) and \( w = -5 \) into the second row: \[ y - \frac{23}{11}(\frac{3}{2}) - \frac{7}{11}(-5) = \frac{43}{11} \] Calculating: \[ y - \frac{69}{22} + \frac{35}{11} = \frac{43}{11} \implies y - \frac{69}{22} + \frac{70}{22} = \frac{43}{11} \implies y + \frac{1}{22} = \frac{86}{22} \implies y = \frac{85}{22} \] Finally, substituting \( y = \frac{85}{22} \), \( z = \frac{3}{2} \), and \( w = -5 \) into the first row: \[ x + 3(\frac{85}{22}) - 5(\frac{3}{2}) - (-5) = 11 \] Calculating: \[ x + \frac{255}{22} - \frac{15}{2} + 5 = 11 \implies x + \frac{255}{22} - \frac{165}{22} + \frac{110}{22} = 11 \implies x + \frac{200}{22} = 11 \implies x = 11 - \frac{200}{22} = \frac{242 - 200}{22} = \frac{42}{22} = \frac{21}{11} \] Thus, the complete solution to the system is: \[ \left( x, y, z, w \right) = \left( \frac{21}{11}, \frac{85}{22}, \frac{3}{2}, -5 \right) \] The correct choice is: A. There is one solution. The solution set is \( \left\{ \left( \frac{21}{11}, \frac{85}{22}, \frac{3}{2}, -5 \right) \right\} \).