Determine whether or not the system of linear equations below can have a unique solution regardless of the value (s) of \( \lambda \). The above mentioned system is given by the augmented matrix below. \( \left[\begin{array}{rr|r}2 & 2 & \lambda \\ -1 & 1 & -3\end{array}\right] \)

To determine whether the system of linear equations represented by the augmented matrix \[ \left[\begin{array}{rr|r}2 & 2 & \lambda \\ -1 & 1 & -3\end{array}\right] \] can have a unique solution regardless of the value of \( \lambda \), we need to analyze the coefficient matrix and the conditions for a unique solution. The coefficient matrix is \[ A = \begin{bmatrix} 2 & 2 \\ -1 & 1 \end{bmatrix} \] A system of linear equations has a unique solution if the coefficient matrix \( A \) is invertible, which occurs when the determinant of \( A \) is non-zero. Let's calculate the determinant of \( A \): \[ \text{det}(A) = (2)(1) - (2)(-1) = 2 + 2 = 4 \] Since the determinant is \( 4 \), which is non-zero, the matrix \( A \) is invertible. This means that the system of equations represented by the augmented matrix will have a unique solution for any value of \( \lambda \). Thus, the answer is: **Yes, the system can have a unique solution regardless of the value of \( \lambda \).**