If $\frac{5}{m}+\frac{2}{n}=9$ and $\frac{3}{m}+\frac{4}{n}=11$ and $mn \neq 0$,then the values of $m$ and $n$ are . . . . . . respectively.

In matrix notation, if the system of equations $\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \\ 10 \end{bmatrix}$ has an infinite number of solutions, then all these solutions lie on

While solving a system of linear equations $AX=B$ using Cramer's rule with the usual notation,if $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5\end{array}\right|$,$\Delta_1=\left|\begin{array}{ccc}5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5\end{array}\right|$ and $X=\left[\begin{array}{l}\alpha \\ 2 \\ \beta\end{array}\right]$,then $\alpha^2+\beta^2=$

Let $S$ be the set of all real values of $k$ for which the system of linear equations $x + y + z = 2$,$2x + y - z = 3$,and $3x + 2y + kz = 4$ has a unique solution. Then $S$ is

If $A=\begin{bmatrix} x & y & y \\ y & x & y \\ y & y & x \end{bmatrix}$ is a matrix such that $5 A^{-1}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$, then $A^2-4 A=$

The system of linear equations $\lambda x + 2y + 2z = 5$,$2\lambda x + 3y + 5z = 8$,and $4x + \lambda y + 6z = 10$ has: