in Linear Algebra edited by
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4 votes
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The condition for which the eigenvalues of the matrix  $A=\begin{bmatrix} 2 & 1\\ 1 &k \end{bmatrix}$   are positive is

  1. $k > \frac{1}{2}$
  2. $ k > −2$
  3. $ k > 0$
  4. $k< \frac{-1}{2}$
in Linear Algebra edited by
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4 Comments

moved by
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$A)K>\frac{1}{2}$
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how?
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1 Answer

1 vote
1 vote

By the properties of eigen values, if all the principal minors of A are possitive then all the eigen values of A are also possitive.

so  |A2*2| > 0

ie 2k - 1 > 0

k>1/2

2 Comments

you are taking det(A)>0..

but what if both eigen values are negative then also det(A)>0 will follow..
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We know that Det(A) = product of eigen values and Tr(A) = sum of eigen values. Therefore, to have both eigen values as positive , det(A) should be >0 & Tr(A) >0.So, we can check for both the conditions on the given matrix:

 

Det(A) > 0 => 2k-1>0 => k>1/2

and Tr(A) > 0 => k+2 > 0 => k>-2

Hence, the first solution satisfies both the condition. So, the answer is A) .. I think this is a better approach than the Best Answer here @Joshi_nitish
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