Question

If A = $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$ and B = $\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$ then the correct relation is
A . ${{A}^{2}}={{B}^{2}}$
B . $A+B=B-A$
C . $AB=BA$
D . None of these

Answer 1

Hint: In this question, we have given two matrices and we have to find out that which option follows. We will solve this question with the help of options. As in the first option, we find the square of A and B then we check whether it will be equal or not. Similarly for second option, we add and subtract the both matrices and check whether it is equal or not and for third option, we multiply both the matrices. By solving all the options, we are able to find out the correct option.

Complete Step- by- step Solution:
Given A = $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$
And B = $\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$
First we check the Option (A)
For this we find the ${{A}^{2}}$ and ${{B}^{2}}$
${{A}^{2}}$= $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$ $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$
${{A}^{2}}$= $\left[ \begin{matrix}
   {{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha & -\cos \alpha \sin \alpha -\sin \alpha \cos \alpha \\
   \sin \alpha \cos \alpha +\cos \alpha \sin \alpha & -{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\
\end{matrix} \right]$
${{A}^{2}}$= $\left[ \begin{matrix}
   {{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha & -2\cos \alpha \sin \alpha \\
   2\sin \alpha \cos \alpha & -{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\
\end{matrix} \right]$
Similarly we find ${{B}^{2}}$
${{B}^{2}}$= $\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$ $\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$
${{B}^{2}}$= $\left[ \begin{matrix}
   {{\cos }^{2}}\beta -{{\sin }^{2}}\beta & -\cos \beta \sin \beta -\cos \beta \sin \beta \\
   \sin \beta \cos \beta +\cos \beta \sin \beta & -{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \\
\end{matrix} \right]$
${{B}^{2}}$= $\left[ \begin{matrix}
   {{\cos }^{2}}\beta -{{\sin }^{2}}\beta & -2\cos \beta \sin \beta \\
   2\sin \beta \cos \beta & -{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \\
\end{matrix} \right]$
Thus ${{A}^{2}}\ne {{B}^{2}}$
Now we check Option (B)
We add A and B, we get
A + B = $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$ + $\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$
A + B = $\left[ \begin{matrix}
   \cos \alpha +\cos \beta & -\sin \alpha -\sin \beta \\
   \sin \alpha +\sin \beta & \cos \alpha +\cos \beta \\
\end{matrix} \right]$
And B – A = $\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$ - $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$
B – A = $\left[ \begin{matrix}
   \cos \beta -\cos \alpha & -\sin \beta +\sin \alpha \\
   \sin \beta -\sin \alpha & \cos \beta -\cos \alpha \\
\end{matrix} \right]$
Thus $A+B\ne B-A$
Now we check Option ( C )
First we multiply A with B
AB = $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$$\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$
AB = $\left[ \begin{matrix}
   \cos \alpha \cos \beta -\sin \alpha \sin \beta & -\cos \alpha \sin \beta -\sin \alpha \cos \beta\\
   sin \alpha \cos \beta +\cos \alpha \sin \beta \ & -\sin \alpha \sin \beta +\cos \alpha \cos \beta \\
\end{matrix} \right]$
Similarly BA = $\left[ \begin{matrix}
   \cos \beta & -\sin \beta \\
   \sin \beta & \cos \beta \\
\end{matrix} \right]$ $\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$
BA = $\left[ \begin{matrix}
   \cos \beta \cos \alpha -\sin \beta \sin \alpha & -\cos \beta \sin \alpha -\sin \beta \cos \alpha \\
   \sin \beta \cos \alpha +\cos \beta \sin \alpha & -\sin \beta \sin \alpha +\cos \beta \cos \alpha \\
\end{matrix} \right]$
Hence $AB=BA$

Thus Option ( C ) is correct.

Note: Students make mistakes in multiplying the matrices. Remember that when we want to do squaring of A, we multiply the matrix A with A and when AB is given then we multiply A with B. If AB is given then first we put matrix A then B and when BA is given then we first put matrix B then A.