GATE Papers >> Mechanical >> 2017 >> Question No 26

Question No. 26 Mechanical | GATE 2017
Consider the matrix $\style{font-family:'Times New Roman'}{\mathrm P=\begin{bmatrix}\frac1{\sqrt2}&0&\frac1{\sqrt2}\\0&1&0\\\frac{-1}{\sqrt2}&0&\frac1{\sqrt2}\end{bmatrix}}$
 
Which one of the following statements about P is INCORRECT?

Answer : (D) All eigenvalues of P are real numbers.


Solution of Question No 26 of GATE 2017 Mechanical Paper

Comsider options one by one

(a) Determinant of P = 1

$ \left|\mathrm P\right|\begin{vmatrix}\frac1{\sqrt2}&0&\frac1{\sqrt2}\\0&1&0\\\frac{-1}2&0&\frac1{\sqrt2}\end{vmatrix}=\frac1{\sqrt2}\left(\frac1{\sqrt2}-0\right)+\frac1{\sqrt2}\left(+\frac1{\sqrt2}\right) $

$ \left|\mathrm P\right|=1 $ option a is correct

$ \begin{array}{l}(\mathrm b)\;\mathrm P^\mathrm T.\mathrm P=\mathrm I=\mathrm{PP}^\mathrm T\\\mathrm P^\mathrm T=\begin{bmatrix}\frac1{\sqrt2}&0&\frac{-1}{\sqrt2}\\0&1&0\\+\frac1{\sqrt2}&0&\frac{\;1}{\sqrt2}\end{bmatrix}\\\mathrm P.\mathrm P^\mathrm T=\begin{bmatrix}\frac1{\sqrt2}&0&\frac1{\sqrt2}\\0&1&0\\\frac{-1}{\sqrt2}&0&\frac{\;1}{\sqrt2}\end{bmatrix}\begin{bmatrix}\frac1{\sqrt2}&0&\frac{-1}{\sqrt2}\\0&1&0\\\frac1{\sqrt2}&0&\frac{\;1}{\sqrt2}\end{bmatrix}\\\mathrm P.\mathrm P^\mathrm T=\begin{bmatrix}\frac12+\frac12&0&-\frac12+\frac12\\0&1&0\\\frac{-1}2+\frac12&0&\frac{\;1}2+\frac12\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\mathrm I\end{array} $

Thus $ \mathrm P.\mathrm P^\mathrm T=\mathrm I\Rightarrow $ option b is correct

(c) as $ \mathrm P.\mathrm P^\mathrm T=\mathrm I $

$ \mathrm P^\mathrm T=\mathrm P^{-1} $

Thus option c is correct

Hence option d is incorrect.

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