Consider the simultaneous linear equations $AX=B$ and $AY=Q$. If $A$ is an invertible matrix and $B$ is the unique solution of $AY=Q$, then the solution of $AX=B$ is

If the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x + y + z = 1$ and $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 9 & 1 \\ 8 & 2 & 1 \\ 7 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$,then $\alpha^2 + \beta^2 + \gamma^2 = $

If ${a_1}x + {b_1}y + {c_1}z = 0, {a_2}x + {b_2}y + {c_2}z = 0, {a_3}x + {b_3}y + {c_3}z = 0$ and $\left| \begin{matrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{matrix} \right| = 0$,then the given system has

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{ll} 1 & 1 \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$, then $A$ is equal to

If $p$ and $q$ are two distinct real values of $\lambda$ for which the system of equations $\begin{aligned} (\lambda-1) x+(3 \lambda+1) y+2 \lambda z &=0 \\ (\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z &=0 \\ 2 x+(3 \lambda+1) y+3(\lambda-1) z &=0 \end{aligned}$ has a non-zero solution, then $p^2+q^2-p q=$

Let $a, b, c, d \in \mathbb{R}$ be such that $ad-bc \neq 0$ and $e$ be a positive number other than $1$. If $x^a y^b=e^m$, $x^c y^d=e^n$, $\Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|$, $\Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$, then the values of $x$ and $y$ are respectively.