Mechanical GATE 2017 Question No 26 Linear Algebra

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?

(A) Determinant of P is equal to 1.
(B) P is orthogonal.
(C) Inverse of P is equal to its transpose.
(D) All eigenvalues of P are real numbers.

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

GATE 2017 Mechanical Linear Algebra Eigen Values and Eigen Vectors

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

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

Thus option c is correct

Hence option d is incorrect.

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

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

Questions of Eigen Values and Eigen Vectors

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