 Solution

Adding three times the second linear equation to the third gives $3(3-\alpha )z=0$. Since $\alpha \ne 3$, it follows that $z=0$. The second linear equation then gives $y=x+1$, and the first becomes $(1+2\alpha )x=1-2\alpha$. Thus the unique solution isThe assumption $\alpha \ne -1/2$ makes the denominator nonzero; $\beta$ is unrestricted.