An ordered pair $(\alpha, \beta)$ for which the system of linear equations $(1 + \alpha)x + \beta y + z = 2$; $\alpha x + (1 + \beta)y + z = 3$; $\alpha x + \beta y + 2z = 2$ has a unique solution,is

If the system of linear equations $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$,where $\alpha, \beta, \gamma \in R$,has infinitely many solutions,then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to

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 $x+4y-z=\lambda$,$7x+9y+\mu z=-3$,and $5x+y+2z=-1$ has infinitely many solutions,then $(2\mu+3\lambda)$ is equal to:

If for $AX = B,$ $B = \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -\frac{1}{2} & -\frac{1}{2} \\ -4 & \frac{3}{4} & \frac{5}{4} \\ 2 & -\frac{1}{4} & -\frac{3}{4} \end{bmatrix},$ then $X$ is equal to

Let $\lambda$ be a real number for which the system of linear equations $x + y + z = 6$,$4x + \lambda y - \lambda z = \lambda - 2$,and $3x + 2y - 4z = -5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation: