Condition for Inconsistency of a Pair of Linear Equations

Video Explanation

Question

For what value of \( \alpha \) will the following system of equations be inconsistent (have no solution)?

\[ \alpha x + 3y = \alpha – 3, \qquad 12x + \alpha y = \alpha \]

Solution

Step 1: Write in Standard Form

\[ \alpha x + 3y – (\alpha – 3) = 0 \quad (1) \]

\[ 12x + \alpha y – \alpha = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = \alpha, \quad b_1 = 3, \quad c_1 = -(\alpha – 3) \]

\[ a_2 = 12, \quad b_2 = \alpha, \quad c_2 = -\alpha \]

Step 3: Condition for Inconsistency

A pair of linear equations is inconsistent if

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Step 4: Apply the Condition

\[ \frac{\alpha}{12} = \frac{3}{\alpha} \]

\[ \alpha^2 = 36 \]

\[ \alpha = \pm 6 \]

Step 5: Check the Third Ratio

For \( \alpha = 6 \):

\[ \frac{c_1}{c_2} = \frac{-3}{-6} = \frac{1}{2}, \qquad \frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2} \]

(All ratios equal ⇒ infinitely many solutions, not acceptable)

For \( \alpha = -6 \):

\[ \frac{a_1}{a_2} = -\frac{1}{2}, \qquad \frac{b_1}{b_2} = -\frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{9}{6} = \frac{3}{2} \]

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Hence, the system is inconsistent.

Conclusion

The given system of equations has no solution for:

\[ \boxed{\alpha = -6} \]

\[ \therefore \quad -6x + 3y = -9 \text{ and } 12x – 6y = -6 \text{ represent parallel lines.} \]