GATE 2014 EC (EC1) – Question 39
$J_6=\begin{pmatrix}0&0&0&0&0&1\\0&0&0&0&1&0\\0&0&0&1&0&0\\0&0&1&0&0&0\\0&1&0&0&0&0\\1&0&0&0&0&0\end{pmatrix}$ reverses the columns of $I_6$. For $P=I_6+\alpha J_6$, $\alpha\ge0$, find $\alpha$ making det(P) = 0.
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Correct answer: 0.99 to 1.01
Explanation
The reversal matrix satisfies J6²=I6. Its symmetric vectors have eigenvalue+1 and antisymmetric vectors have eigenvalue−1, three of each. Thus det(I6+αJ6)=(1+α)³(1−α)³. The roots are±1; with α nonnegative the answer is **1**.