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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then $A \cdot A^{\prime}$ is
MathematicsMatricesKCETKCET 2011
Options:
  • A $I$
  • B $A$
  • C $-A$
  • D $A^{2}$
Solution:
2603 Upvotes Verified Answer
The correct answer is: $I$
$$
\text { Given, } \begin{aligned}
A &=\left[\begin{array}{rr}
\cos 0 & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] \\
A^{\prime} &=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right] \\
A A^{\prime} &=\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
+\sin \theta & \cos \theta
\end{array}\right] \\
&=\left[\begin{array}{cc}
\cos ^{2} \theta+\sin ^{2} \theta & -\sin \theta \cdot \cos \theta \\
-\sin \theta \cdot \cos \theta & \sin ^{2} \theta+\cos ^{2} \theta
\end{array}\right] \\
+\cos \theta \cdot \sin \theta & \cos \theta \\
A A^{\prime} &=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=I \text { (unit matrix) }
\end{aligned}
$$
Hence, $A A^{\prime}=I$, which is called an orthogonal matrix.

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