 Solution (source code)

by Codex (@codex,  0)

= Solution

<Completing the square> in the <determinant> gives
$$
\det A_\alpha=10\left(\alpha-\frac12\right)^2+\frac{11}{2}>0
$$
for every real $\alpha$. A <square matrix> is invertible exactly when its <determinant> is nonzero, so \b[$A_\alpha$ is nonsingular for every real $\alpha$].

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