Let and be the distinct roots of the equation x2+x-1=0. Consider the set T=\1, , \. For a 3 3 matrix M= (aij )3 3, define Ri=ai1+ai2+ai3 and Cj=a1j+a2j+a3j for i=1,2,3 and j=1,2,3. Match each entry in List-I to the correct entry in List-II.
- (P) → (4) (Q) → (2) (R) → (5) (S) → (1)
- (P) → (2) (Q) → (4) (R) → (1) (S) → (5)
- (P) → (2) (Q) → (4) (R) → (3) (S) → (5)
- (P) → (1) (Q) → (5) (R) → (3) (S) → (4)
Answer
(3) (P) → (2) (Q) → (4) (R) → (3) (S) → (5)
Solution
+ =-1 1+ + =0. With 0.618, -1.618, the only way three entries of T add to 0 is \1, , \ (each once).
(P) Every row and column is a permutation of 1, , : Latin squares of order 3 =12. P → 2
(Q) Symmetric with column sums 0: each symbol occurs once on the diagonal (off-diagonal entries come in pairs); diagonal in 3!=6 ways fixes the rest, so count =6. Q → 4
(R) M j=(a12,0,-a23)T, so a solution exists; a 3 3 skew-symmetric matrix is singular, hence infinitely many solutions. R → 3
(S) M(1,1,1)T=0 |M|=0. S → 5
Answer: (C)