For which value(s) of \(\lambda\) do the pair of linear equations \(\lambda x + y = \lambda^2\) and \(x + \lambda y = 1\) have (i) no solution, (ii) infinitely many solutions, (iii) a unique solution?
(i) \(\lambda = -1\); (ii) \(\lambda = 1\); (iii) \(\lambda \neq \pm 1\).
Write both in standard form.
\(\lambda x + y - \lambda^2 = 0\)
\(x + \lambda y - 1 = 0\)
Here \(a_1 = \lambda,\ b_1 = 1,\ c_1 = -\lambda^2\).
And \(a_2 = 1,\ b_2 = \lambda,\ c_2 = -1\).
Unique solution if the determinant is non-zero.
\(\Delta = a_1 b_2 - a_2 b_1 = \lambda^2 - 1\).
So, unique solution when \(\lambda^2 - 1 \neq 0\).
That is \(\lambda \neq \pm 1\).
Parallel (no solution) when
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}.\)
Compute: \(\dfrac{a_1}{a_2} = \lambda\), \(\dfrac{b_1}{b_2} = \dfrac{1}{\lambda}\).
Equality gives \(\lambda = \dfrac{1}{\lambda}\Rightarrow \lambda^2=1\).
At \(\lambda = -1\), the constants ratio is
\(\dfrac{c_1}{c_2} = \dfrac{-\lambda^2}{-1} = \lambda^2 = 1\).
The first two ratios are \(-1\) but the third is \(1\).
Hence, lines are parallel ⇒ no solution.
Coincident (infinitely many) when all three ratios are equal.
At \(\lambda = 1\), we have \(\dfrac{a_1}{a_2} = 1\), \(\dfrac{b_1}{b_2} = 1\),
and \(\dfrac{c_1}{c_2} = 1\).
So the lines coincide ⇒ infinitely many solutions.