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Matrices & Determinants: Let Pmatrix Pmatrix Pmatrix Pmatrix Identity Matrix Order

JEE Maths question with a full step-by-step solution.

Question
Let A=(2070101−21)A = \begin{pmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{pmatrix} and B=(−α14α7α010α−4α−2α)B = \begin{pmatrix}-\alpha & 14\alpha & 7\alpha\\ 0 & 1 & 0\\ \alpha & -4\alpha & -2\alpha\end{pmatrix}. If AB=IAB = I, where II is the identity matrix of order 3, then the trace of BB equals:
A00
B25\dfrac{2}{5}correct
C15\dfrac{1}{5}
D55
Solution
Step 1: Determine α\alpha from the (1,1)(1,1) entry of ABAB
(AB)11=2(−α)+0(0)+7(α)=5α(AB)_{11} = 2(-\alpha) + 0(0) + 7(\alpha) = 5\alpha
 For AB=IAB = I: 5α=1  ⟹  α=155\alpha = 1 \implies \alpha = \dfrac{1}{5}. Step 2: Compute the trace
tr(B)=(−α)+1+(−2α)=1−3α=1−35=25\text{tr}(B) = (-\alpha) + 1 + (-2\alpha) = 1 - 3\alpha = 1 - \frac{3}{5} = \frac{2}{5}
 Answer: (2)
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