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

$A$ trust fund has Rs. $30,000$ that must be invested in two different types of bonds. The first bond pays $5 \%$ interest per year,and the second bond pays $7 \%$ interest per year. Using matrix multiplication,determine how to divide Rs. $30,000$ among the two types of bonds if the trust fund must obtain an annual total interest of Rs. $1800$.

Let $n$ be the number obtained on rolling a fair die. If the probability that the system of equations
$x-ny+z=6$
$x+(n-2)y+(n+1)z=8$
$(n-1)y+z=1$
has a unique solution is $\frac{k}{6}$, then the sum of $k$ and all possible values of $n$ is:

If the system of equations $x+2y+3z=3$,$4x+3y-4z=4$,and $8x+4y-\lambda z=9+\mu$ has infinitely many solutions,then the ordered pair $(\lambda, \mu)$ is equal to

The number of solutions of the equations $x + y - z = 0$,$3x - y - z = 0$,and $x - 3y + z = 0$ is

If the system of equations $x+ky+3z=-2$, $4x+3y+kz=14$, and $2x+y+2z=3$ can be solved by the matrix inversion method, then: