$A, C$ are $3 \times 3$ matrices. $B, D$ are $3 \times 1$ matrices. If $AX=B$ has a unique solution and $CX=D$ has an infinite number of solutions, then:

If a system of three linear equations in three unknowns,which is in the matrix equation form of $AX = D$,is inconsistent,then $\frac{\text{rank of } A}{\text{rank of } AD}$ is

If the system of equations,$a^2 x - ay = 1 - a$ and $bx + (3 - 2b) y = 3 + a$ possess a unique solution $x = 1, y = 1$,then:

Statement $-1$: The system of linear equations
$x + (\sin \alpha)y + (\cos \alpha)z = 0$
$x + (\cos \alpha)y + (\sin \alpha)z = 0$
$x - (\sin \alpha)y - (\cos \alpha)z = 0$
has a non-trivial solution for only one value of $\alpha$ lying in the interval $(0, \frac{\pi}{2})$.
Statement $-2$: The equation in $\alpha$
$\left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} \right| = 0$
has only one solution lying in the interval $(0, \frac{\pi}{2})$.

The system of equations $\begin{cases} \alpha x + y + z = \alpha - 1 \\ x + \alpha y + z = \alpha - 1 \\ x + y + \alpha z = \alpha - 1 \end{cases}$ has no solution,if $\alpha$ is

Consider the system of equations: $x + ay = 0$,$y + az = 0$,and $z + ax = 0$. The set of all real values of $a$ for which the system has a unique solution is: