Question

For what values of k is the system of equations kx + 3y = k – 2, 12x + ky = k inconsistent?

Answer 1

Given: kx + 3y = k – 2, 12x + ky = k

Step-wise calculation:

1. The pair has no solution exactly when the two rows of coefficients are proportional but the constants are not i.e., `(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`.

2. Compute determinant of coefficient matrix: det = k² – 36. If det ≠ 0 there is a unique solution; for possible inconsistency we need det = 0

⇒ k² – 36 = 0 

⇒ k = 6 or k = –6

3. Check k = 6: 

Equations become 6x + 3y = 4 and 12x + 6y = 6.

Divide the first by 3: `2x + y = 4/3`; divide the second by 6: 2x + y = 1.

These give `2x + y = 4/3` and 2x + y = 1, a contradiction ⇒ inconsistent for k = 6.

4. Check k = –6:

Equations become –6x + 3y = –8 and 12x – 6y = –6.

Divide the first by 3: `-2x + y = -8/3`; divide the second by 6: 2x – y = –1.

Adding them gives `0 = -11/3`, a contradiction ⇒ inconsistent for k = –6.

The system is inconsistent exactly for k = 6 and k = –6.