Let P1 = I = [1&0&0 0&1&0 0&0&1], P2 = [1&0&0 0&0&1 0&1&0], P3 = [0&1&0 1&0&0 0&0&1], P4 = [0&1&0 0&0&1 1&0&0], P5 = [0&0&1 1&0&0 0&1&0], P6 = [0&0&1 0&1&0 1&0&0] and x = displaystyle∑6K = 1PK[2&1&3 1&0&2 3&2&1]PTK where PTK denotes the transpose of the matrix PK. Then which of the following options is/are correct?

Question

Let P1 = I = [1&0&0 0&1&0 0&0&1], P2 = [1&0&0 0&0&1 0&1&0], P3 = [0&1&0 1&0&0 0&0&1], P4 = [0&1&0 0&0&1 1&0&0], P5 = [0&0&1 1&0&0 0&1&0], P6 = [0&0&1 0&1&0 1&0&0] and x = displaystyle∑6K = 1PK[2&1&3 1&0&2 3&2&1]PTK where PTK denotes the transpose of the matrix PK. Then which of the following options is/are correct? (A) If X[1 1 1] = α[1 1 1], then α = 30 (B) X is a symmetric matrix (C) The sum of diagonal entries of X is 18 (D) X — 30I is an invertible matrix

Tardigrade Admin · Accepted Answer

Let Q=[2&1&3 1&0&2 3&2&1] X= displaystyle ∑k=16 (PkQPTK) XT= = displaystyle ∑k=16 (PkQPTK)T=X. X is symmetric Let R=[1 1 1] XR= displaystyle ∑k=16PKQPKTR.[∵ PK TR=R] = displaystyle ∑K=16PK=[2&2&2 2&2&2 2&2&2] QR=[6 3 6] ⇒ XR=[2&2&2 2&2&2 2&2&2][6 3 6]=[30 30 30]=30R ⇒ α =30. Trace X = Trace ( displaystyle ∑K=16PKQPKT) = displaystyle ∑K=16 Trace (PKQPKT)=6 (TraceQ) =18 X[1 1 1]=30[1 1 1] ⇒ (X-30 I)[1 1 1]=O ⇒ |X-30 I|=0 ⇒ X-30 I is non-invertible

If X⎣⎡​111​⎦⎤​=α⎣⎡​111​⎦⎤​, then α = 30

X is a symmetric matrix

The sum of diagonal entries of X is 18

X — 30I is an invertible matrix

If $X ⎣ ⎡ 111 ⎦ ⎤ = α ⎣ ⎡ 111 ⎦ ⎤,$ then $α$ = 30