Let A = (ai,j_{i,j}i,j ) be a real symmetric 3 × 3 matrix. Consider the quadratic formQ(x1,x2,x3)=xTAx,where x=(x1,x2,x3)TQ(x_1, x_2, x_3) = \m
Question
Let A = (ai,j ) be a real symmetric 3 × 3 matrix. Consider the quadratic form
Q(x1,x2,x3)=xTAx,where x=(x1,x2,x3)T.
Which of the following is true?
A.
If Q(x1,x2,x3) is positive definite, then ai,j > 0 for all i = j.
B.
If Q(x1,x2,x3) is positive definite, then ai,i > 0 for all i.
C.
If ai,j > 0 for all i = j, then Q(x1,x2,x3) is positive definite.
D.
. If ai,i > 0 for all i, then Q(x1,x2,x3) is positive definite.
Correct option is B
LetA=adedbfefc,and supposeQ(x1,x2,x3) is positive definite.Positive Definite Condition:For Q(x1,x2,x3) to be positive definite:1. All principal minors of A must be positive:D1=[a],D2=[addb],D3=adedbfefc.2. Calculating the determinants:D1=a>0.D2=det[addb]=ab−d2>0⟹ab>d2.D3=det(A)=abc+2def−af2−be2−cd2>0.Verification of Options:Option A:If Q(x1,x2,x3) is positive definite, this does not imply that ai,j>0 for i=j.For example, consider A=42−3251−316:abc+2def=108,af2+be2+cd2=73,108>73,but e=−3<0. Thus, ai,j>0 for i=j is not necessary for Q to be positive definite.Option A is incorrect.Option B:If Q(x1,x2,x3) is positive definite, then all diagonal elements ai,i>0.From the positivity of principal minors:D1=a>0,D2=ab−d2>0⟹b>0,D3>0⟹c>0.Hence, ai,i>0 for all i.Option B is correct.Option C:If ai,j>0 for i=j, this does not guarantee that Q(x1,x2,x3) is positive definite.For example, if A has positive off-diagonal entries but fails to satisfy the positivity of principal minors, Q will not be positive definite.Option C is incorrect.Option D:If ai,i>0 for all i, this does not guarantee that Q(x1,x2,x3) is positive definite.For instance, if A=467651716,where a,b,c>0, but if the determinant D3≤0,Q will not be positive definite.Option D is incorrect.Final Answer:Option B is correct.
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