1
Question
If P=∣∣
∣∣√3212−12√32∣∣
∣∣, A=[1101] and Q=PAPT, then PTQ50P is
If P=∣∣
∣∣√3212−12√32∣∣
∣∣, A=[1101] and Q=PAPT, then PTQ50P is
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Solution
The correct option is A [15001]
Given:P=⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦, A=[1101]
Here, PPT=⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦⎡⎢
⎢
⎢⎣√32−1212√32⎤⎥
⎥
⎥⎦=[1001]=I
Also, PTP=⎡⎢
⎢
⎢⎣√32−1212√32⎤⎥
⎥
⎥⎦⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦=[1001]=I
⇒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(Q50)P
=P−1(Q.Q49)P=(AP−1)Q49Pby(1)
=AP−1(Q.Q48)P
=A(AP−1)Q48P=A2P−1Q48P
Repeatedly applying (1) , ultimately we get
X=A50P−1P⇒A50=[15001]
Hence, option 'A' is correct.
Given:
P=⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦, A=[1101]
Here,
Here,
PPT=⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦⎡⎢
⎢
⎢⎣√32−1212√32⎤⎥
⎥
⎥⎦=[1001]=I
Also,
Also,
PTP=⎡⎢
⎢
⎢⎣√32−1212√32⎤⎥
⎥
⎥⎦⎡⎢
⎢
⎢⎣√3212−12√32⎤⎥
⎥
⎥⎦=[1001]=I
⇒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(Q50)P
=P−1(Q.Q49)P=(AP−1)Q49Pby(1)
=AP−1(Q.Q48)P
=A(AP−1)Q48P=A2P−1Q48P
Repeatedly applying (1) , ultimately we get
X=A50P−1P⇒A50=[15001]
A3=[1301] and so on...
Now, Q=PAPT=PAP−1
⇒P−1Q=AP−1 ......(1)
Let X=PT(Q50)P
=P−1(Q.Q49)P=(AP−1)Q49Pby(1)
=AP−1(Q.Q48)P
=A(AP−1)Q48P=A2P−1Q48P
Repeatedly applying (1) , ultimately we get
X=A50P−1P⇒A50=[15001]
Hence, option 'A' is correct.
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