. For what values of λ ∈ R, the following system of equations has (i) no solution, (ii)a unique solution, and (iii) infinitely many solutions?(5 − λ)x + 4y + 2z = 4, 4x + (5 − λ)y + 2z = 4, 2x + 2y + (2 − λ)z = 2.Also find the solutions whenever they exist

The given system of equations can be written in matrix form as follows:

A*X = B, where

A = [[5-λ, 4, 2], [4, 5-λ, 2], [2, 2, 2-λ]], X = [x, y, z]^T, and B = [4, 4, 2]^T.

(i) The system has no solution if and only if the matrix A is singular (i.e., det(A) = 0) and the vector B is not in the column space of A. The determinant of A is given by:

det(A) = -(λ-3)^2 * (λ-4).

Setting this equal to zero gives the roots λ = 3, 4. However, for these values of λ, B is in the column space of A, so the system has either a unique solution or infinitely many solutions, not no solution.

(ii) The system has a unique solution if and only if the matrix A is non-singular (i.e., det(A) ≠ 0). From the above, we see that det(A) ≠ 0 for λ ≠ 3, 4. So, the system has a unique solution for all λ ∈ R except λ = 3, 4.

(iii) The system has infinitely many solutions if and only if the matrix A is singular (i.e., det(A) = 0) and the vector B is in the column space of A. From the above, we see that this occurs for λ = 3, 4.

For λ = 3, the system of equations becomes:

2x + 4y + 2z = 4, 4x + 2y + 2z = 4, 2x + 2y + z = 2.

Subtracting the third equation from the first gives z = 0. Substituting z = 0 into the third equation gives x + y = 1. So, the solutions are of the form (x, 1-x, 0), where x ∈ R.

For λ = 4, the system of equations becomes:

x + 4y + 2z = 4, 4x + y + 2z = 4, 2x + 2y + z = 2.

Subtracting the third equation from the first gives x = 0. Substituting x = 0 into the third equation gives y + z = 1. So, the solutions are of the form (0, y, 1-y), where y ∈ R.