Find the matrix $X$ such that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.

The system of equations $kx + y + z = 1$,$x + ky + z = k$,and $x + y + kz = k^2$ has no solution if $k$ is equal to

Let $S$ be the set of all column matrices $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ such that $b_1, b_2, b_3 \in \mathbb{R}$ and the system of equations (in real variables)
$-x+2y+5z=b_1$
$2x-4y+3z=b_2$
$x-2y+2z=b_3$
has at least one solution. Then,which of the following system$(s)$ (in real variables) has (have) at least one solution for each $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] \in S$?
$(A)$ $x+2y+3z=b_1, 4y+5z=b_2$ and $x+2y+6z=b_3$
$(B)$ $x+y+3z=b_1, 5x+2y+6z=b_2$ and $-2x-y-3z=b_3$
$(C)$ $-x+2y-5z=b_1, 2x-4y+10z=b_2$ and $x-2y+5z=b_3$
$(D)$ $x+2y+5z=b_1, 2x+3z=b_2$ and $x+4y-5z=b_3$

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 linear equations $x + ky + 3z = 0$,$3x + ky - 2z = 0$,and $2x + 4y - 3z = 0$ has a non-zero solution $(x, y, z)$,then $\frac{xz}{y^2} = \dots$

If the system of simultaneous linear equations $x+y+z=\lambda$,$5x-y+\mu z=10$,and $2x+3y-z=6$ has a unique solution,then: