a) \( \left\{\begin{array}{l}(\lambda-1) x+2 y-2=0 \\ (\lambda-2) x+(\lambda-1) y=1\end{array}\right. \)
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To solve the system of equations \[ \begin{aligned} (\lambda-1)x + 2y &= 2 \quad \quad (1) \\ (\lambda-2)x + (\lambda-1)y &= 1 \quad \quad (2) \end{aligned} \] we'll represent it in matrix form and consider the determinant to discuss the conditions for consistent solutions. The determinant can become zero for certain values of \( \lambda \), which may indicate no solutions or infinitely many solutions. By setting up the augmented matrix and finding the determinant, we get: \[ \begin{bmatrix} \lambda - 1 & 2 \\ \lambda - 2 & \lambda - 1 \end{bmatrix} \] The determinant is given by: \[ D = (\lambda - 1)(\lambda - 1) - 2(\lambda - 2) = \lambda^2 - 2\lambda + 1 - 2\lambda + 4 = \lambda^2 - 4\lambda + 5 \] This quadratic does not necessarily simplify to a zero but can be analyzed for roots. For the system to have a unique solution, we need \( D \neq 0 \). For \( D = 0 \), the discriminant \( (-4)^2 - 4 \times 1 \times 5 = 16 - 20 = -4 \) shows that this polynomial will always be positive, confirming that there are no values of \( \lambda \) where the determinant equals zero. Thus, there are infinitely many solutions. Substituting values of \( \lambda \) would allow further exploration of particular solutions. For \( \lambda = 1 \) and \( \lambda = 2 \) specifically, we obtain: 1. For \( \lambda = 1 \): The system becomes inconsistent if tried directly by substitution. 2. For \( \lambda = 2 \): The equations collapse to a single line due to dependability. Thus there are infinitely many solutions. Hence, solving for all potential values of \( \lambda \) confirms a unique result every time, guiding further analysis in each special case.
