Question

If we have a matrix as If \[A\left( \alpha \right) = \left[ {\begin{array}{*{20}{c}}
  {\cos \alpha }&{\sin \alpha } \\
 { - \sin \alpha }&{\cos \alpha }
\end{array}} \right]\], then the matrix \[{A^2}\left( \alpha \right) = \]
\[1){\text{ }}A\left( {2\alpha } \right)\]
\[2){\text{ }}A\left( \alpha \right)\]
\[3){\text{ }}A\left( {3\alpha } \right)\]
\[4){\text{ }}A\left( {4\alpha } \right)\]

Answer 1

Hint: We have to find the value of the matrix \[{A^2}\left( \alpha \right)\] . We solve this using the concept of operations of matrices . We should have the knowledge of the cases for which the multiplication of two matrices is possible or not . For solving this problem we should also have the knowledge of the various trigonometric identities for the double angle of sine function , cosine function . We should also have the knowledge about the order of a matrix .

Complete step-by-step solution:
Given : \[A\left( \alpha \right) = \left[ {\begin{array}{*{20}{c}}
  {\cos \alpha }&{\sin \alpha } \\
  { - \sin \alpha }&{\cos \alpha }
\end{array}} \right]\]
The order of \[A\left( \alpha \right)\]is \[2 \times 2\]

Multiplication of matrices
Let \[A = \left[ {\begin{array}{*{20}{c}}
  a&b
\end{array}} \right]\]and \[B = \left[ {\begin{array}{*{20}{c}}
  c \\
  d
\end{array}} \right]\]
Then , the product of AB is given as :
\[AB = \left[ {a \times c + b \times d} \right]\]
For multiplication the number of columns of the first matrix should be equal to the number of rows of the second matrix .
Using multiplication of matrices , we get
\[{A^2}\left( \alpha \right) = \left[ {\begin{array}{*{20}{c}}
  {\cos \alpha }&{\sin \alpha } \\
  { - \sin \alpha }&{\cos \alpha }
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  {\cos \alpha }&{\sin \alpha } \\
  { - \sin \alpha }&{\cos \alpha }
\end{array}} \right]\]
\[{A^2}\left( \alpha \right) = \left[ {\begin{array}{*{20}{c}}
  {{{\cos }^2}\alpha - {{\sin }^2}\alpha }&{\sin \alpha \cos \alpha + \cos \alpha \sin \alpha } \\
  { - (\sin \alpha \cos \alpha + \cos \alpha \sin \alpha )}&{{{\cos }^2}\alpha - {{\sin }^2}\alpha }
\end{array}} \right]\]
We know that ,
Double angle formula of sin function and cos function is given as :
\[\sin 2x = 2\sin x\cos x\]
\[\cos 2x = {\cos ^2}x - {\sin ^2}x\]
Using the double angle formula , we get
${A^2}(\alpha ) = \left( {\begin{array}{*{20}{c}}
  {cos2\alpha }&{sin2\alpha } \\
  { - sin2\alpha }&{cos2\alpha }
\end{array}} \right)$
From above relation , we get
${A^2}(\alpha ) = A(2\alpha )$
Thus , the value of \[{A^2}\left( \alpha \right)\]is \[A\left( {2\alpha } \right)\]
Hence , the correct option is \[\left( 1 \right)\].

Note: The properties of multiplication of matrices :
(1) Associative Law :
For any three matrices \[A\] , \[B\] and \[C\] . We have\[\left( {AB} \right)C = A\left( {BC} \right)\], whenever both sides of the equality are defined .
(2) Distributive Law :
For any three matrices \[A\] , \[B\] and \[C\].
\[\left( i \right)\] \[A\left( {B + C} \right) = AB + AC\] , \[(ii)\] \[\left( {A + B} \right)C = AC + BC\] whenever both sides of equality are defined .
(3) The existence of multiplicative identity :
For every square matrix \[A\], there exists an identity matrix of the same order such that \[IA = AI = A\].