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Suppose that the trace of a 2×2 matrix A is tr(A)=3, and the determinant is det(A)=−40. Find the eigenvalues of A
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I think you need to solve below : \(\large \lambda^2 - \lambda (\text{trace} ) + \text{determinant} = 0 \)
\[ \begin{bmatrix} a&b\\ c&d \end{bmatrix} \]
\[ (a-\lambda)(d-\lambda)-bc=0 \]We know \(ad-bc = -40\) and \(a+d=3\).
Make that \(\lambda_1\) and \(\lambda_2\).
\[ ad-(a+d)\lambda+\lambda^2-bc = 0 \]Then we get: \[ \lambda^2-(\color{red}{a+d})\lambda +(\color{blue}{ad-bc})=0 \]So looking at that, I agree with @ganeshie8
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Thank you! I got my small eigenvalue and large eigenvalues mixed up. small eigen.= -5, and large eigen. = 8
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