If the system of linear equations $x + y + z = 1$, $x + 2y + 4z = \eta$, $x + 4y + 10z = \eta^2$ has a solution, then the value of $\eta$ is:
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Look for linear dependency in the columns or rows: $3 \times (\text{Eq. 2}) - 2 \times (\text{Eq. 1}) = (\text{Eq. 3})$ holds for the LHS coefficients. Hence, the same relationship must hold for the RHS: $3\eta - 2 = \eta^2 \implies \eta^2 - 3\eta + 2 = 0$.