Question

For what value of α, the system of equations

αx + 3y = α - 3

12x + αy =  α

will have no solution?

Answer 1

The given system of the equation may be written as

αx + 3y -α - 3 = 0

12x + αy - α = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = α, b_1 = 3, c_1 = -(α - 3)`

And `a_2 = 12, b_2 = α, c_2 = -α`

For a unique solution, we must have

`a_1/a_2 - b_1/b_2 != c_1/c_2`

`=> α/12 = 3/α != (-(α - 3))/(-α)`

Now,

`3/α != (-(α - 3))/(-α)`

`=> 3/α != (α - 3)/α`

`=> 3 != α - 3`

`=> 3 = 3 != α`

`=> 6 != α`

`=> α != 6`

And

`α/12 = 3/α`

`=> α^2 = 36`

`=> α +- 6`

`=> α = -6`       [∵ `α != 6`]

Hence, the given system of equation will have no solution if α = -6