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Question

Let
P1=∣ ∣100010001∣ ∣, P2=∣ ∣100001010∣ ∣, P3=∣ ∣010100001∣ ∣, P4=∣ ∣010001100∣ ∣, P5=∣ ∣001100010∣ ∣, P6=∣ ∣001010100∣ ∣, and X=6k=1Pk213102321PTk, where PTk denotes the transpose of the matrix Pk. Then which of the following options is/are correct?

A
X30I is an invertible matrix
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B
The sum of diagonal entries of X is 18
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C
X is a symmetric matrix
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D
If X111=α111, then α=30
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Solution

The correct option is D If X111=α111, then α=30
Let A=213102321
X=6k=1(PkAPTk)=(P1APT1+P2APT2++P6APT6)
XT=6k=1(PkAPTk)T=X [A=AT]
X is a symmetric matrix.

Let Q=111

PT1Q=100010001111=111=Q,

Similarly, PTkQ=Q
XQ=6k=1PkAPTkQ
=6k=1PkAQ
=(6k=1Pk)AQ
6k=1Pk=222222222 and AQ=636
XQ=222222222636=303030
XQ=30111=30Q=αQ
α=30
Also, X111=30111(X30I)111=0
(X30I)Q=0
The system has a non-trivial solution.
So, |X30I|=0X30I is a non-invertible matrix.

Trace(X)=Trace(6k=1PkAPTk)=6 Trace(A)=6×3=18[Trace(A)=2+0+1=3]

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