= Solution
Subtracting the first equation from the third gives $4x-4z=6$, so $z=x-3/2$. The second equation then becomes $(\alpha-2)x=\beta-3$. The third fixes $y=7(1-x)/5$. Therefore the three possibilities are
$$
\boxed{\begin{array}{c|c}
\alpha\ne2 & \text{one solution},\quad
x=\dfrac{\beta-3}{\alpha-2},\ y=\dfrac75(1-x),\ z=x-\dfrac32\\[4pt]
\alpha=2,\ \beta=3 & \text{infinitely many},\quad
(x,y,z)=\left(t,\dfrac75(1-t),t-\dfrac32\right)\\[4pt]
\alpha=2,\ \beta\ne3 & \text{no solution}.
\end{array}}
$$
For the first case the coefficient <determinant> is $20(\alpha-2)\ne0$, confirming uniqueness. At $\alpha=2$ its <matrix rank> is two. The <augmented matrix> has the same <matrix rank> only for $\beta=3$, giving one <free variable in a linear system>; otherwise the <linear system> is inconsistent.
Back to article page
