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+3 votes
The condition for which the eigenvalues of the matrix  $A=\begin{bmatrix} 2 & 1\\ 1 &k \end{bmatrix}$   are positive is
$(A)  k > \frac{1}{2}$

$(B)  k > −2$

$(C)  k > 0$

$(D)  k< \frac{-1}{2}$
asked in Linear Algebra by Boss (27.1k points)
edited by | 551 views
matrix = 21...what it means??

1 Answer

0 votes

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


answered by Boss (12k points)
you are taking det(A)>0..

but what if both eigen values are negative then also det(A)>0 will follow..
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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