Let P 1 = I = 1 0 0 0 1 0 0 0 1 ,   P 2 = 1 0 0 0 0 1 0 1 0 ,   P 3 = 0 1 0 1 0 0 0 0 1 ,  P…

Let P1=I=100010001, P2=100001010, P3=010100001, P4=010001100, P5=001100010,P6=001010100, and
X=∑k=16PK213102321PKT
where PKT denotes the transpose of the matrix PK. Then which of the following options is/are correct?
  1. X-30I is an invertible matrix
  2. The sum of diagonal entries of X is 18
  3. If X111=α111, then α=30
  4. X is a symmetric matrix

Solution

∵ P1=P1T=P1-1,P3=P3T=P3-1,P5=P5T=P5T,P2=P2T=P2-1,P4=P4T=P4T,P6=P6T=P6-1
Now
Let Q=213102321
X=∑K=16PKQPKT
⇒TraceX=∑K=16TrPKQPKT∵TrAB=TrBA
=∑K=16TrPKTPKQ∵PKT PK=I
=∑K=16TrQ
=6 TraceQ
=6×3
=18
⇒X=∑K=16PKQPKT
XT=∑K=16PKQPKTT
XT=∑K=16PKTTQTPKT∵ATT=A
XT=∑K=16PKQPKT∵QT=Q
XT=X
⇒ Let R=111 then XR=∑K=16PKQPKTR
XR=∑K=16PKQPKR∵PKT=PK
XR=∑K=16PKQR∵PKR=R
XR=∑K=16PK636∵QR=636
XR=222222222636∵ ∑K=16PK=222222222
XR=303030⇒XR=30111⇒α=30
⇒XR=30111
XR=30 R
X-30IR=0
Now x-30I must be zero else R=0
If x-30I≠0
then x-30I-1x-30IR=0
R=0 not possible
Hence, x-30I=0
X-30I is not invertible.

Asked in: JEE Advanced 2019 (Paper 2)

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