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If A=[a1cb2121c]A=\begin{bmatrix}a & 1 & c\\ b & 2 & 1\\ 2 & 1 & c\end{bmatrix} is a singular matrix, then

Solution

✅ Correct Option: 2

Expanding, ∣A∣=a(2c−1)−(bc−2)+c(b−4)=2ac−a−4c+2=(2c−1)(a−2)|A|=a(2c-1)-(bc-2)+c(b-4)=2ac-a-4c+2=(2c-1)(a-2).

Singularity means (2c−1)(a−2)=0(2c-1)(a-2)=0, and bb never appears, so bb is free.

Note: The determinant vanishes when a=2a=2 or c=12c=\frac{1}{2}; option 2 states one such pair with bb free and is taken as the intended key.

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