Question

Consider the matrix P = $\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ \frac{-1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$ 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 eigen values of P are real numbers

Answer 1

The correct option is D All eigen values of P are real numbers

P = $\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$

(i) |P| = 1

(ii) $P^T$ = $\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{-1}{\sqrt{2}}\\ 0& 1& 0\\ \frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$

P.$P^T$ = $\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ \frac{-1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$ $\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{-1}{\sqrt{2}}\\ 0& 1& 0\\ \frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$

= $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ = I

Hence P is orthongonal as P.$P^T$ = I

(iii) $P^{-1}$ = $\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{-1}{\sqrt{2}}\\ 0& 1& 0\\ \frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$ = $P^T$

Hence (iv) is wrong.