If $$A = \left[ {\matrix{ {5a} & { - b} \cr 3 & 2 \cr } } \right]$$ and $$A$$ adj $$A=A$$ $${A^T},$$ then $$5a+b$$ is equal to :

The set of all values of $$\lambda $$ for which the system of linear equations:

$$\matrix{ {2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}} \cr {2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr { - {x_1} + 2{x_2} = \lambda {x_3}} \cr } $$

has a non-trivial solution

If $$A = \left[ {\matrix{ 1 & 2 & 2 \cr 2 & 1 & { - 2} \cr a & 2 & b \cr } } \right]$$ is a matrix satisfying the equation

$$A{A^T} = 9\text{I},$$ where $$I$$ is $$3 \times 3$$ identity matrix, then the ordered

pair $$(a, b)$$ is equal to :

If $$\alpha ,\beta \ne 0,$$ and $$f\left( n \right) = {\alpha ^n} + {\beta ^n}$$ and $$$\left| {\matrix{ 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr } } \right|$$$
$$ = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},$$ then $$K$$ is equal to :