Consider the following system of linear equations.x + 2y − z = 32x + 5y + z = 7x + y − (k + 1)2z = −k1Determine all values of constant k for which the following system of linear equations has(a) no solutions(b) unique solution(c) infinitely many solutions(d) For k = 2, find the solution for x using Cramer’s rule

(a) No Solutions: For a system of linear equations to have no solutions, the system must be inconsistent. This happens when the coefficients of x, y, and z in two equations are proportional, but the constants on the right side of the equation are not.

Let's compare the first two equations:

x + 2y - z = 3 and 2x + 5y + z = 7

If we multiply the first equation by 2, we get:

2x + 4y - 2z = 6

Comparing this with the second equation, we see that the coefficients of x and y are not proportional (4y is not equal to 5y), so the system is consistent for all values of k. Therefore, there are no values of k for which the system has no solutions.

(b) Unique Solution: For a system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.

The coefficient matrix is:

1 2 -1 2 5 1 1 1 -(k+1)^2

The determinant of this matrix is:

1*(5*(-(k+1)^2) - 1) - 2*(2*(-(k+1)^2) - 1) + 1*(2 - 5)

Solving this equation for k, we find that the determinant is non-zero for all k ≠ 2. Therefore, the system has a unique solution for all k ≠ 2.

(c) Infinitely Many Solutions: For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero.

From the previous part, we know that the determinant is zero when k = 2. Therefore, the system has infinitely many solutions when k = 2.

(d) For k = 2, find the solution for x using Cramer’s rule: Cramer's rule states that the solution for x can be found by replacing the first column of the coefficient matrix with the constants on the right side of the equations, and then dividing by the determinant of the original coefficient matrix.

The modified matrix is:

3 2 -1 7 5 1 -2 1 -(3)^2

The determinant of this matrix is:

3*(5*(-9) - 1) - 2*(7*(-9) - (-2)) - 1*(7 - 14) = -120

The determinant of the original coefficient matrix when k = 2 is 0.

Therefore, the solution for x when k = 2 is -120 / 0, which is undefined. This is consistent with our previous finding that the system has infinitely many solutions when k = 2.