The system of equations

$$\begin{aligned} & x+y+z=6, \\ & x+2 y+5 z=9, \\ & x+5 y+\lambda z=\mu, \end{aligned}$$

has no solution if

Let $A=\left[a_{i j}\right]$ be a $3 \times 3$ matrix such that $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], A\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$ and $A\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$, then $a_{23}$ equals :

If the system of equations

$$ \begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\ & (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9 \end{aligned}$$

has infinitely many solutions, then $\lambda^2+\lambda$ is equal to

If $\mathrm{A}, \mathrm{B}, \operatorname{and}\left(\operatorname{adj}\left(\mathrm{A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)$ are non-singular matrices of same order, then the inverse of $A\left(\operatorname{adj}\left(A^{-1}\right)+\operatorname{adj}\left(B^{-1}\right)\right)^{-1} B$, is equal to