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Question

If A=[0tanα2tanα20] and I is the identity matrix of order 2, show that I + A =(IA)[cosαsinαsinαcosα]


Solution

Here, A=[0tt0], where t=tan(α2)
Now, cosα=1tan2(α2)1+tan2(α2)=1t21+t2 and sinα=2tan(α2)1+tan2(α2)=2t1+t2
RHS=(IA)[cosαsinαsinαcosα]=[(1001)(0t+t0)]1t21+t22t1+t22t1+t21t21+t2

=[1tt1]1t21+t22t1+t22t1+21t21+t2=⎢ ⎢1t2+2t21+t22t+t(1t2)1+t2t(1t2)+2t1+t22t2+1t21+t2⎥ ⎥=1+t21+t22t+tt31+t2t+t3+2t1+22t2+1t21+t2=⎢ ⎢1+t21+t2t(1+t2)1+t2t(1+t2)1+t21+t21+t2⎥ ⎥=[1tt1]LHS=[1001]+[0tt0]=[0+1t+0t+00+1]=[1tt1]=RHS

Putting the value of t on both sides, we get
[1tan(α2)tan(α2)1]=[1tan(α2)tan(α2)1]
LHS=RHS. hence proved.

Alternale method Here, A=[0tan(α2)tan(α2)0]
I+A=[1001]+[0tan(α2)tan(α2)0]=[1tan(α2)tan(α2)0]
and IA=[1001][0tan(α2)tan(α2)0]=[1tan(α2)tan(α2)1]

(IA)[cosαsinαsinαcosα]=[1tan(α2)tan(α2)1][cosαsinαsinαcosα]=[cosα+tan(α2)sinαsinα+tan(α2)cosαtan(α2)cosα+sinαtan(α2)sinα+cosα]=⎢ ⎢ ⎢ ⎢cosαcos(α2)+sin(α2)sinθcos(α2)sinαcos(α2)+sin(α2)cosαcos(α2)sin(α2)cosα+cos(α2)sinαcos(α2)sin(α2)sinα+cos(α2)cosαcos(α2)⎥ ⎥ ⎥ ⎥

=⎢ ⎢ ⎢ ⎢cos(αα2)cos(α2)sin(αα2)cos(α2)sin(αα2)cos(α2)cos(αα2)cos(α2)⎥ ⎥ ⎥ ⎥=⎢ ⎢ ⎢ ⎢cos(α2)cos(α2)sin(α2)cos(α2)sin(α2)cos(α2)cos(α2)cos(α2)⎥ ⎥ ⎥ ⎥( cos(AB)=cos A cos B+sin A sin Bsin(AB)=sin A cos Bcos A sin B)=[1tan(α2)tan(α2)1]=I+A
Thus, I+A=(IA)[cosαsinαsinαcosα] Hence proved.


Mathematics

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