Equality of Matrices

For infinitely many solutions, the third equation must be a linear combination of the first two.

Let (iii) = $$\alpha$$(i) + $$\beta$$(ii). Matching coefficients:

$$2\alpha + 5\beta = 100$$ ... (A), $$-\alpha + \lambda\beta = -47$$ ... (B), $$\alpha + 3\beta = \mu$$ ... (C), $$4\alpha + 12\beta = 212$$ ... (D)

From (D): $$\alpha + 3\beta = 53$$, so $$\mu = 53$$.

From (A) and $$\alpha = 53 - 3\beta$$: $$106 - 6\beta + 5\beta = 100$$, giving $$\beta = 6$$ and $$\alpha = 35$$.

From (B): $$-35 + 6\lambda = -47$$, so $$\lambda = -2$$.

$$\mu - 2\lambda = 53 - 2(-2) = 57$$.

The correct answer is Option 1: 57.