Question

Assertion :If A is an orthogonal matrix of order 2, then $|A|=\pm 1$. Reason: Every two-rowed real orthogonal matrix is of any one of the forms $\left(\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right)$ or $\left(\begin{array}{cc}cos\theta & sin\theta \\ sin\theta & -cos\theta \end{array}\right)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Let $A$ be $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\because A$ is orthogonal $A{A}^T=I$

$\therefore \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]=I\Rightarrow \left[\begin{array}{cc}{a}^2+b^2& ac+bc\\ ac+bd& c^2+d^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\therefore a^2+b^2=1c^2+d^2=1ac+bd=0$

Let $\frac{a}{d}=\frac{-b}{c}=k------\left(1\right)\therefore c^2+d^2=\frac{1}{k^2}\therefore k^2=1\Rightarrow k=\pm 1$

So $\frac{a}{d}=\frac{-b}{c}=\pm 1\therefore (a,b,c,d)\in [-1,1]$ (From(1))

Let $a=cos\theta ,b=sin\theta [\because cos\theta \&sin\theta$ lies between $0$ & $1]$

We know $\frac{a}{d}=\frac{-b}{c}=-1$

So $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

Now for $\frac{a}{d}=\frac{-b}{c}=-1,\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$