(a) \[ \left[\begin{array}{lll:l} 1 & 0 & 4 & 4 \\ 0 & 1 & 3 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right] \] The system has no solution. The system has a unique solution. \[ (x, y, z)=\square, \square, \square \] The system has infinitely many solutions. \[ \begin{array}{l} (x, y, z)=(x, \square, \square \\ (x, y, z)=\square, y, \square \\ (x, y, z)=\square, \square, z) \end{array} \]

Step 1 ：The given matrix is in row-echelon form. The last row of the matrix represents the equation 0x + 0y + 0z = 0, which is always true for any values of x, y, and z.

Step 2 ：The second row represents the equation y + 3z = 3, and the first row represents the equation x + 4z = 4.

Step 3 ：Since there is no row in the matrix that represents an equation that is always false (like 0 = 1), the system of equations has solutions.

Step 4 ：Since the third column has a leading 1 (the number 1 is the first non-zero number from the top in its column) in the second row but not in the first or third rows, the variable z is a free variable. This means that z can be any real number, and there are infinitely many solutions to the system of equations.

Step 5 ：We can express x and y in terms of z using the first two rows of the matrix. From the first row, we have x = 4 - 4z, and from the second row, we have y = 3 - 3z.

Step 6 ：Final Answer: The system has infinitely many solutions. The solutions can be expressed as: \(\boxed{(x, y, z) = (4 - 4z, 3 - 3z, z)}\) for any real number z.