Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right)$$. Then A²⁰²⁵ $$-$$ A²⁰²⁰ is equal to :

Let $$\theta \in \left( {0,{\pi \over 2}} \right)$$. If the system of linear equations

$$(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0$$

$${\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0$$

$${\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0$$

has a non-trivial solution, then the value of $$\theta$$ is :

If $$A = \left( {\matrix{ {{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr {{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr } } \right)$$, $$B = \left( {\matrix{ 1 & 0 \cr i & 1 \cr } } \right)$$, $$i = \sqrt { - 1} $$, and Q = A^TBA, then the inverse of the matrix A Q²⁰²¹ A^T is equal to :

Let A and B be two 3 $$\times$$ 3 real matrices such that (A² $$-$$ B²) is invertible matrix. If A⁵ = B⁵ and A³B² = A²B³, then the value of the determinant of the matrix A³ + B³ is equal to :