1
Question
If P=⎡⎢⎣√3212−12√32⎤⎥⎦ and A=[1101] and Q=PAPT and x=PTQ2005P, then X is equal to
If P=⎡⎢⎣√3212−12√32⎤⎥⎦ and A=[1101] and Q=PAPT and x=PTQ2005P, then X is equal to
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Solution
The correct option is A [1200501]
Given P=⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦,A=[1101]
Here, PPT=⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦⎡⎢
⎢
⎢⎣√32−1212√32⎤⎥
⎥
⎥⎦=[1001]
Also, PTP=⎡⎢
⎢
⎢⎣√32−1212√32⎤⎥
⎥
⎥⎦⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦=[1001]
⇒PTP=PPT=I
i.e. P is orthogonal.
⇒PT=P−1
Also , A2=[1101][1101]=[1201]
A3=[1301] and so on .
Now,Q=PAPT=PAP−1
⇒P−1Q=AP−1 ......(1)
Let X=PT(Q2005)P
=P−1(Q.Q2004)P=(AP−1)Q2004Pby(1)
=AP−1(Q.Q2003)P
=A(AP−1)Q2003P=A2P−1Q2003P
Repeatedly applying (1) , we get ultimately
X=A2005P−1P⇒X=A2005=[1200501]
Given P=⎡⎢ ⎢ ⎢⎣√3212−12√32⎤⎥ ⎥ ⎥⎦,A=[1101]
Here, PPT=⎡⎢ ⎢ ⎢⎣√3212−12√32⎤⎥ ⎥ ⎥⎦⎡⎢ ⎢ ⎢⎣√32−1212√32⎤⎥ ⎥ ⎥⎦=[1001]
Also, PTP=⎡⎢ ⎢ ⎢⎣√32−1212√32⎤⎥ ⎥ ⎥⎦⎡⎢ ⎢ ⎢⎣√3212−12√32⎤⎥ ⎥ ⎥⎦=[1001]
⇒PTP=PPT=I
i.e. P is orthogonal.
⇒PT=P−1
Also , A2=[1101][1101]=[1201]
A3=[1301] and so on .
Now,Q=PAPT=PAP−1
⇒P−1Q=AP−1 ......(1)
Let X=PT(Q2005)P
=P−1(Q.Q2004)P=(AP−1)Q2004Pby(1)
=AP−1(Q.Q2003)P
=A(AP−1)Q2003P=A2P−1Q2003P
Repeatedly applying (1) , we get ultimately
X=A2005P−1P⇒X=A2005=[1200501]
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