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Maths-

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Easy

Question

Let a matrix A = $open square brackets table row 1 1 row 0 1 end table close square brackets$ & P = $open square brackets table row cell fraction numerator square root of 3 over denominator 2 end fraction end cell cell fraction numerator 1 over denominator 2 end fraction end cell row cell negative fraction numerator 1 over denominator 2 end fraction end cell cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table close square brackets$ Q = PAP^T where P^T is transpose of matrix P. Find PT Q2005 P is

$[\begin{matrix} 1 & 2005 \\ 0 & 1 \end{matrix}]$
$\frac{1}{4} [\begin{matrix} 1 + 2005 \sqrt{3} & 6015 \\ 2005 & 1 - 2005 \sqrt{3} \end{matrix}]$
$\frac{1}{4} [\begin{matrix} 1 + 2005 \sqrt{3} & 2005 \\ 2005 & 1 - 2005 \sqrt{3} \end{matrix}]$
$[\begin{matrix} 2005 & 2005 \\ 0 & 1 \end{matrix}]$

The correct answer is: $open square brackets table row 1 2005 row 0 1 end table close square brackets$

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