Suppose the vectors x₁, x₂ and x₃ are the
solutions of the system of linear equations,
Ax = b when the vector b on the right side is equal to b₁, b₂ and b₃ respectively. if

$${x_1} = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$, $${x_2} = \left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right]$$, $${x_3} = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right]$$

$${b_1} = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$$, $${b_2} = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$$ and $${b_3} = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$$,
then the determinant of A is equal to :

If the system of equations
x+y+z=2
2x+4y–z=6
3x+2y+$$\lambda $$z=$$\mu $$
has infinitely many solutions, then

If $$A = \left[ {\matrix{ {\cos \theta } & {i\sin \theta } \cr {i\sin \theta } & {\cos \theta } \cr } } \right]$$, $$\left( {\theta = {\pi \over {24}}} \right)$$

and $${A^5} = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$, where $$i = \sqrt { - 1} $$ then which one of the following is not true?

Let A be a 3 $$ \times $$ 3 matrix such that
adj A = $$\left[ {\matrix{ 2 & { - 1} & 1 \cr { - 1} & 0 & 2 \cr 1 & { - 2} & { - 1} \cr } } \right]$$ and B = adj(adj A).

If |A| = $$\lambda $$ and |(B^-1)^T| = $$\mu $$ , then the ordered pair,
(|$$\lambda $$|, $$\mu $$) is equal to :