Question

Найдите, при каком отрицательном значении параметра k система: 2х + ky = 5, kx + 2y = 6 не имеет решений.

Answer 1

2x+ky=5
kx+2y=6
Суммируем эти уравнения:
2x+kx+ky+2y=11
x(2+k)+y(2+k)=11
(2+k)(x+y)=11
2+k≠0
k≠-2
Ответ:  k≠-2.

Answer 2

Finding the Negative Value of Parameter k for the System to Have No Solutions

To find the negative value of the parameter k for which the system of equations 2x + ky = 5 and kx + 2y = 6 has no solutions, we can use the determinant of the coefficient matrix.

The coefficient matrix for the system is:

| 2   k |
| k   2 |

The determinant of this matrix is given by:

D = (2 * 2) - (k * k) = 4 - k^2

For the system to have no solutions, the determinant D must be equal to zero. Therefore, we can set up the equation:

4 - k^2 = 0

Solving this equation will give us the negative value of the parameter k for which the system has no solutions.

Let's solve the equation:

4 - k^2 = 0

Subtracting 4 from both sides:

-k^2 = -4

Dividing both sides by -1:

k^2 = 4

Taking the square root of both sides:

k = ±2

Since we are looking for the negative value of k, the solution is:

k = -2

Therefore, when the parameter k is equal to -2, the system of equations 2x + ky = 5 and kx + 2y = 6 has no solutions.

Note: The solution provided above is based on the mathematical calculation of the determinant of the coefficient matrix. However, it is always a good practice to verify the solution by substituting the value of k back into the original system of equations and checking if it leads to a contradiction or inconsistency.