An ordered pair ($$\alpha $$, $$\beta $$) for which the system of linear equations
(1 + $$\alpha $$) x + $$\beta $$y + z = 2
$$\alpha $$x + (1 + $$\beta $$)y + z = 3
$$\alpha $$x + $$\beta $$y + 2z = 2
has a unique solution, is :

If $$\lambda $$ be the ratio of the roots of the quadratic equation in x, 3m²x² + m(m – 4)x + 2 = 0, then the least value of m for which $$\lambda + {1 \over \lambda } = 1,$$ is

Let P = $$\left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 0 \cr 9 & 3 & 1 \cr } } \right]$$ and Q = [q_ij] be two 3 $$ \times $$ 3 matrices such that Q – P⁵ = I₃.

Then $${{{q_{21}} + {q_{31}}} \over {{q_{32}}}}$$ is equal to :

Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by n_i, the label of the ball drawn from the i^th box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n₁ < n₂ < n₃ is :