The number of values of $$k$$, for which the system of equations : $$$\matrix{ {\left( {k + 1} \right)x + 8y = 4k} \cr {kx + \left( {k + 3} \right)y = 3k - 1} \cr } $$$
has no solution, is

Let $$P$$ and $$Q$$ be $$3 \times 3$$ matrices $$P \ne Q.$$ If $${P^3} = {Q^3}$$ and
$${P^2}Q = {Q^2}P$$ then determinant of $$\left( {{P^2} + {Q^2}} \right)$$ is equal to :

Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right)$$. If $${u_1}$$ and $${u_2}$$ are column matrices such
that $$A{u_1} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right)$$ and $$A{u_2} = \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right),$$ then $${u_1} + {u_2}$$ is equal to :

The number of values of $$k$$ for which the linear equations
$$4x + ky + 2z = 0,kx + 4y + z = 0$$ and $$2x+2y+z=0$$ possess a non-zero solution is :