The set of values of $k$ for which the system of simultaneous equations $x+y+kz=1$,$2x+2y=3$,and $x+2y+2kz=k$ has no real solution 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:

The positive value of $a$ for which the system of linear homogeneous equations $x+ay+z=0$, $ax+2y-z=0$, and $2x+3y+z=0$ has non-trivial solutions is

Let $A=\begin{bmatrix} 0 \\ -6 \\ 8 \end{bmatrix}$,$B=\begin{bmatrix} 3 & 5 & -7 \\ 0 & -1 & 8 \\ 6 & -1 & 0 \end{bmatrix}$ and $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$. If $D=[\alpha, \beta, \gamma]^{T}$ is the solution of $X^{T} B^{T}=A^{T}$,then $D^{T} A=$

If the system of equations $x+y+z=5$, $x+2y+2z=6$, and $x+3y+\lambda z=\mu$ (where $\lambda, \mu \in R$) is solvable by the Matrix Inversion Method, then:

For the system $S$ of linear equations $x+y+z=3, 2x+2y-z=3, x+y+\lambda z=1$, the incorrect option among the following statements is: