Let R= pmatrixx&0&0\\0&y&0\\0&0&z pmatrix be a non-zero 3 3 matrix, where x =y ( + 2 3 )=z ( + 4 3 ) 0, (0,2 ). For a square matrix M, let trace (M) denote the sum of all the diagonal entries of M. Then, among the statements: (I) Trace (R)=0 (II) If trace (adj(adj(R)))=0, then R has exactly one non-zero entry.
- Only (I) is true
- Only (II) is true
- Both (I) and (II) are true
- Neither (I) nor (II) is true
Answer
(2) Only (II) is true
Solution
Let x =y ( + 2 3 )=z ( + 4 3 )=k 0 x,y,z 0
∵ + ( + 2 3 )+ ( + 4 3 )=0: kx+ ky+ kz=0 xy+yz+zx=0
(I): if also x+y+z=0, then
x2+y2+z2=(x+y+z)2-2(xy+yz+zx)=0
x=y=z=0, impossible
trace (R)=x+y+z 0 (I) is false
(II): adj(adj R)=|R|3-2R=xyz R
trace =xyz(x+y+z) 0
∵ the condition 'trace =0' never holds, the statement 'if … then …' is (vacuously) true (II) is true
Answer: (2)
Note: The official NTA answer key gives option (4). Our faculty team has re-checked this question and finds option (2) correct, as explained above.