To solve the system of equations using Gaussian elimination, we first represent the system as an augmented matrix. Then, we perform row operations to transform the matrix into row-echelon form. From there, we continue to reduce it to reduced row-echelon form if possible. Finally, we interpret the resulting matrix to determine if there is a unique solution, infinitely many solutions, or no solution.
To solve the given system of equations using Gaussian elimination, we will first write the system in augmented matrix form and then perform row operations to reach row-echelon form. Finally, we will interpret the results to find the solution.
The given system of equations is:
\[
\begin{array}{r}
w + 3x + 4y - z = 7 \\
3x - 4y + z = 5 \\
w - 6x + y = 3
\end{array}
\]
The augmented matrix for this system is:
\[
\begin{bmatrix}
1 & 3 & 4 & -1 & | & 7 \\
0 & 3 & -4 & 1 & | & 5 \\
1 & -6 & 1 & 0 & | & 3
\end{bmatrix}
\]
First, we will eliminate the \(w\) term from the third row by subtracting the first row from the third row:
\[
R_3 = R_3 - R_1
\]
\[
\begin{bmatrix}
1 & 3 & 4 & -1 & | & 7 \\
0 & 3 & -4 & 1 & | & 5 \\
0 & -9 & -3 & 1 & | & -4
\end{bmatrix}
\]
Next, we will eliminate the \(x\) term from the second row by dividing the second row by 3:
\[
R_2 = \frac{1}{3}R_2
\]
\[
\begin{bmatrix}
1 & 3 & 4 & -1 & | & 7 \\
0 & 1 & -\frac{4}{3} & \frac{1}{3} & | & \frac{5}{3} \\
0 & -9 & -3 & 1 & | & -4
\end{bmatrix}
\]
Now, eliminate the \(x\) term from the third row by adding 9 times the second row to the third row:
\[
R_3 = R_3 + 9R_2
\]
\[
\begin{bmatrix}
1 & 3 & 4 & -1 & | & 7 \\
0 & 1 & -\frac{4}{3} & \frac{1}{3} & | & \frac{5}{3} \\
0 & 0 & -15 & 4 & | & 11
\end{bmatrix}
\]
Now, we will perform back substitution to find the values of the variables. From the third row, we have:
\[
-15y + 4z = 11 \quad \Rightarrow \quad y = \frac{4z - 11}{15}
\]
From the second row, we have:
\[
x - \frac{4}{3}y + \frac{1}{3}z = \frac{5}{3}
\]
Substitute \(y = \frac{4z - 11}{15}\) into the equation:
\[
x - \frac{4}{3}\left(\frac{4z - 11}{15}\right) + \frac{1}{3}z = \frac{5}{3}
\]
Simplify:
\[
x - \frac{16z - 44}{45} + \frac{1}{3}z = \frac{5}{3}
\]
Multiply through by 45 to clear the fractions:
\[
45x - (16z - 44) + 15z = 75
\]
\[
45x - 16z + 44 + 15z = 75
\]
\[
45x - z = 31
\]
\[
x = \frac{31 + z}{45}
\]
From the first row, we have:
\[
w + 3x + 4y - z = 7
\]
Substitute \(x = \frac{31 + z}{45}\) and \(y = \frac{4z - 11}{15}\):
\[
w + 3\left(\frac{31 + z}{45}\right) + 4\left(\frac{4z - 11}{15}\right) - z = 7
\]
Simplify:
\[
w + \frac{93 + 3z}{45} + \frac{16z - 44}{15} - z = 7
\]
Multiply through by 45 to clear the fractions:
\[
45w + 93 + 3z + 3(16z - 44) - 45z = 315
\]
\[
45w + 93 + 48z - 132 - 45z = 315
\]
\[
45w + 3z - 39 = 315
\]
\[
45w + 3z = 354
\]
\[
w = \frac{354 - 3z}{45}
\]
The solution set is:
\[
\{(w, x, y, z) = \left(\frac{354 - 3z}{45}, \frac{31 + z}{45}, \frac{4z - 11}{15}, z\right)\}
\]
where \(z\) is any real number. Therefore, the system has infinitely many solutions.
\[
\boxed{\text{B. There are infinitely many solutions.}}
\]
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