(a) No solutions:

For a system of linear equations to have no solutions, the system must be inconsistent. This happens when the coefficients of the variables form a plane and the constants form a line that does not intersect the plane.

We can check this by comparing the ratios of the coefficients of the variables to the constants.

For the first and second equations, the ratios are:

1/(3+2k) = 1/(5+12k)

Solving this equation gives k = -3/2.

For the first and third equations, the ratios are:

1/(3+2k) = 1/(2k)

Solving this equation gives k = 1.

Since the values of k are different, the system is inconsistent and there are no solutions when k = -3/2 or k = 1.

(b) Unique solution:

For a system of linear equations to have a unique solution, the determinant of the coefficients of the variables must not be zero.

The determinant of the coefficients is:

1*4*k^2 - 1*1*k^2 - 1*4*1 + 1*1*1 - k^2*1*1 = 0

Solving this equation gives k = -1 or k = 2.

So, the system has a unique solution when k = -1 or k = 2.

(c) Infinitely many solutions:

For a system of linear equations to have infinitely many solutions, the system must be dependent. This happens when the ratios of the coefficients of the variables to the constants are the same for all equations.

From the previous parts, we know that the system is dependent when k = -3/2 or k = 1.

(d) For k = 1, find a solution for x of the above system using Cramer’s rule:

Cramer's rule states that the solution for a variable is the determinant of the matrix formed by replacing the coefficients of that variable with the constants, divided by the determinant of the coefficients.

The determinant of the coefficients is:

1*4*1 - 1*1*1 - 1*4*1 + 1*1*1 - 1*1*1 = 0

The determinant of the matrix formed by replacing the coefficients of x with the constants is:

(3+2)*4*1 - (3+2)*1*1 - (5+12)*4*1 + (5+12)*1*1 - 2*1*1 = -16

So, the solution for x when k = 1 is -16/0, which is undefined. Therefore, there is no solution for x when k = 1.

Question

(a) No solutions:

For a system of linear equations to have no solutions, the system must be inconsistent. This happens when the coefficients of the variables form a plane and the constants form a line that does not intersect the plane.

We can check this by comparing the ratios of the coefficients of the variables to the constants.

For the first and second equations, the ratios are:

1/(3+2k) = 1/(5+12k)

Solving this equation gives k = -3/2.

For the first and third equations, the ratios are:

1/(3+2k) = 1/(2k)

Solving this equation gives k = 1.

Since the values of k are different, the system is inconsistent and there are no solutions when k = -3/2 or k = 1.

(b) Unique solution:

For a system of linear equations to have a unique solution, the determinant of the coefficients of the variables must not be zero.

The determinant of the coefficients is:

1*4*k^2 - 1*1*k^2 - 1*4*1 + 1*1*1 - k^2*1*1 = 0

Solving this equation gives k = -1 or k = 2.

So, the system has a unique solution when k = -1 or k = 2.

(c) Infinitely many solutions:

For a system of linear equations to have infinitely many solutions, the system must be dependent. This happens when the ratios of the coefficients of the variables to the constants are the same for all equations.

From the previous parts, we know that the system is dependent when k = -3/2 or k = 1.

(d) For k = 1, find a solution for x of the above system using Cramer’s rule:

Cramer's rule states that the solution for a variable is the determinant of the matrix formed by replacing the coefficients of that variable with the constants, divided by the determinant of the coefficients.

The determinant of the coefficients is:

1*4*1 - 1*1*1 - 1*4*1 + 1*1*1 - 1*1*1 = 0

The determinant of the matrix formed by replacing the coefficients of x with the constants is:

(3+2)*4*1 - (3+2)*1*1 - (5+12)*4*1 + (5+12)*1*1 - 2*1*1 = -16

So, the solution for x when k = 1 is -16/0, which is undefined. Therefore, there is no solution for x when k = 1.

Knowee AI · Accepted Answer

(a) No solutions:

For a system of linear equations to have no solutions, the system must be inconsistent. This happens when the coefficients of the variables form a plane and the constants form a line that does not intersect the plane.

We can check this by comparing the ratios of the coefficients of the variables to the constants.

For the first and second equations, the ratios are:

1/(3+2k) = 1/(5+12k)

Solving this equation gives k = -3/2.

For the first and third equations, the ratios are:

1/(3+2k) = 1/(2k)

Solving this equation gives k = 1.

Since the values of k are different, the system is inconsistent and there are no solutions when k = -3/2 or k = 1.

(b) Unique solution:

For a system of linear equations to have a unique solution, the determinant of the coefficients of the variables must not be zero.

The determinant of the coefficients is:

1*4*k^2 - 1*1*k^2 - 1*4*1 + 1*1*1 - k^2*1*1 = 0

Solving this equation gives k = -1 or k = 2.

So, the system has a unique solution when k = -1 or k = 2.

(c) Infinitely many solutions:

For a system of linear equations to have infinitely many solutions, the system must be dependent. This happens when the ratios of the coefficients of the variables to the constants are the same for all equations.

From the previous parts, we know that the system is dependent when k = -3/2 or k = 1.

(d) For k = 1, find a solution for x of the above system using Cramer’s rule:

Cramer's rule states that the solution for a variable is the determinant of the matrix formed by replacing the coefficients of that variable with the constants, divided by the determinant of the coefficients.

The determinant of the coefficients is:

1*4*1 - 1*1*1 - 1*4*1 + 1*1*1 - 1*1*1 = 0

The determinant of the matrix formed by replacing the coefficients of x with the constants is:

(3+2)*4*1 - (3+2)*1*1 - (5+12)*4*1 + (5+12)*1*1 - 2*1*1 = -16

So, the solution for x when k = 1 is -16/0, which is undefined. Therefore, there is no solution for x when k = 1.

Question

Solution

Similar Questions

Upgrade your grade with Knowee