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Find the eigenvalues of the matrix {\begin{bmatrix}5&2\\3&6\\\end{bmatrix}}.

First, make sure that the determinant is not zero.

5\cdot 6-2\cdot 3=24\,

Now subtract \lambda \, from each entry of the main diagonal.

{\begin{bmatrix}5-\lambda &2\\3&6-\lambda \\\end{bmatrix}}

Build and solve the characteristic equation the same way as for the determinant. Set it equal to zero.

(5-\lambda )\cdot (6-\lambda )-6=0\,

(30-11\lambda +\lambda ^{2})-6=0\,

\lambda ^{2}-11\lambda +24=0\,

The quadratic formula (or factoring) tells the answer:

\lambda ={\frac  {11\pm {\sqrt  {121-4\cdot 1\cdot (24)}}}{2}}\,

\lambda =3,8\,

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