Final answer:
To have a unique solution, the value of k cannot be 8. For infinitely many solutions, h must be 3, and k must be 8.
Step-by-step explanation:
The question involves finding the conditions for a unique solution or infinitely many solutions to a system of linear equations. To determine the possible values of h and k for which the system
x₁ + 2x₂ = h
4x₁ + kx₂ = 12
has a unique solution, we use the concept of non-singularity of the coefficient matrix. If the determinant of the coefficient matrix is not zero, the system has a unique solution. In this case, the determinant is
(1)(k) - (2)(4) = k - 8.
For a unique solution, the determinant must be non-zero:
k - 8 ≠ 0
k ≠ 8
Therefore, for any value of h and k ≠ 8, the system will have a unique solution.
For infinitely many solutions, the system must be dependent, meaning that one equation is a scalar multiple of the other. This happens when the ratios of the coefficients and the constants are equal, leading to
(1/4) = (2/k) = (h/12)
From the first two ratios, we get k = 8. Substituting k = 8 into the third ratio, we get h = 3. Thus, the system will have infinitely many solutions when h = 3 and k = 8.