The equations $x-y+2z=4$,$3x+y+4z=6$,and $x+y+z=1$ have

If the system of linear equations $x+y+3z=0$,$x+3y+k^{2}z=0$,and $3x+y+3z=0$ has a non-zero solution $(x, y, z)$ for some $k \in R$,then $x + (y/z)$ is equal to

Let $k_1$ and $k_2$ be the maximum and minimum values of $k$ for which the system of equations $x + ky = 1$,$kx + y = 2$,and $x + y = k$ are consistent. Then $k_1^2 + k_2^2$ is equal to:

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be a $2 \times 2$ real matrix with $\det(A) = 1$. If the equation $\det(A - \lambda I_2) = 0$ has imaginary roots (where $I_2$ is the identity matrix of order $2$), then:

The value of $k \in R$,for which the system of linear equations
$3x - y + 4z = 3$
$x + 2y - 3z = -2$
$6x + 5y + kz = -3$
has infinitely many solutions,is:

Let $A=\begin{bmatrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{bmatrix}$ and $B=[b_{ij}]_{3 \times 3}$ with $b_{11}=2, b_{13}=-2, b_{12}=0$ such that $AB=\begin{bmatrix} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{bmatrix}$. Then $|B|+\operatorname{trace}(B)=$