For what value of $k$ does the following system of equations possess a non-trivial solution?
$x + ky + 3z = 0$
$3x + ky - 2z = 0$
$2x + 3y - 4z = 0$

Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4$,$3x+2y+5z=3$,$9x+4y+(28+[\lambda])z=[\lambda]$ has a solution is:

If the system of equations $kx + (k+1)y + (k-1)z = 0$,$(k-1)x + (k+2)y + kz = 0$,and $(k+1)x + ky + (k+2)z = 0$ has a non-trivial solution,then the sum of all possible values of $k$ is:

Solve the system of linear equations using the matrix method: $2x - y = -2$ and $3x + 4y = 3$.

The following system of equations $3x - 7y + 5z = 3$,$3x + y + 5z = 7$,and $2x + 3y + 5z = 5$ is:

Let $\alpha, \beta (\alpha \neq \beta)$ be the values of $m$ for which the equations $x+y+z=1$,$x+2y+4z=m$,and $x+4y+10z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}(n^\alpha+n^\beta)$ is equal to: