If \(A=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\), then \(A \cdot A^{\prime}\) is

\[
\text { Given, } \begin{aligned}
A &=\left[\begin{array}{rr}
\cos 0 & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] \\
A^{\prime} &=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right] \\
A A^{\prime} &=\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
+\sin \theta & \cos \theta
\end{array}\right] \\
&=\left[\begin{array}{cc}
\cos ^{2} \theta+\sin ^{2} \theta & -\sin \theta \cdot \cos \theta \\
-\sin \theta \cdot \cos \theta & \sin ^{2} \theta+\cos ^{2} \theta
\end{array}\right] \\
+\cos \theta \cdot \sin \theta & \cos \theta \\
A A^{\prime} &=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=I \text { (unit matrix) }
\end{aligned}
\]
Hence, \(A A^{\prime}=I\), which is called an orthogonal matrix.