The solution of $$\det (A - \lambda {I_2}) = 0$$ be 4 and 8 and $$A = \left( {\matrix{ 2 & 2 \cr x & y \cr } } \right)$$. Then

(I₂ is identity matrix of order 2)

If M is a 3 $$\times$$ 3 matrix such that (0, 1, 2) M = (1 0 0), (3, 4 5) M = (0, 1, 0), then (6 7 8) M is equal to

Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 0 & {\cos t} & {\sin t} \cr 0 & { - \sin t} & {\cos t} \cr } } \right)$$

Let $$\lambda$$₁, $$\lambda$$₂, $$\lambda$$₃ be the roots of $$\det (A - \lambda {I_3}) = 0$$, where I₃ denotes the identity matrix. If $$\lambda$$₁ + $$\lambda$$₂ + $$\lambda$$₃ = $$\sqrt 2 $$ + 1, then the set of possible values of t, $$-$$ $$\pi$$ $$\ge$$ t < $$\pi$$ is

Let A and B two non singular skew symmetric matrices such that AB = BA, then A²B²(A^TB)^$$-$$1(AB^$$-$$1)^T is equal to