+3 votes
591 views

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. $k< \frac{-1}{2}$

edited | 591 views
0
matrix = 21...what it means??
0

for the eigen values to be positive here i think...

k+2>0 and its determinant >0

k>-2 and 2k-1>0 ---------> k>1/2

so to satisfy both the cases domain will be  k>1/2.

solution will be option A.

+1
0
$A)K>\frac{1}{2}$
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how?

## 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

k>1/2

by Boss (11.9k points)
0
you are taking det(A)>0..

but what if both eigen values are negative then also det(A)>0 will follow..
+7
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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