Let $A$ be a $3\times 3$ real matrix such that $A\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right);A\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)$ and $A\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)$. If $X={\left[x_1 x_2 x_3\right]}^T$ and $I$ is an identity matrix of order $3$, then the system $\left(A-2I\right)X=\left(\begin{array}{l}4\\ 1\\ 1\end{array}\right)$ has

The system of equations
$-kx+3y-14z=25$

$-15x+4y-kz=3$

$-4x+y+3z=4$

Question: is consistent for all $k$ in the set

Let $X=\left[\begin{array}{lll}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right],Y=\alpha I+\beta X+\gamma X^2$ and $Z={\alpha }^{2}I-\alpha \beta X+\left({\beta }^2-\alpha \gamma \right)X^2,\alpha ,\beta ,\gamma \in \mathrm{ℝ}$.

If ${\mathrm{Y}}^{-1}=\left[\begin{array}{ccc}\frac{1}{5}& \frac{-2}{5}& \frac{1}{5}\\ 0& \frac{1}{5}& \frac{-2}{5}\\ 0& 0& \frac{1}{5}\end{array}\right]$, then ${\left(\alpha -\beta +\gamma \right)}^2$ is equal to ______.

The ordered pair $\left(a,b\right)$, for which the system of linear equations
$3x-2y+z=b$

$5x-8y+9z=3$

$2x+y+az=-1$
has no solution, is