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Recall that the characteristic polynomial of a 2x2 matrix [tex]\mathbf A=\begin{bmatrix}a&b\\c&d\end{bmatrix}[/tex] is

[tex]\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}a-\lambda&b\\c&d-\lambda\end{vmatrix}=(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)[/tex]

but [tex]\det(\mathbf A)=ad-bc[/tex] and [tex]\mathrm{tr}(\mathbf A)=a+d[/tex], so the characteristic polynomial for [tex]\mathbf A[/tex] is

[tex]\lambda^2-\mathrm{tr}(\mathbf A)\lambda+\det(\mathbf A)[/tex]

We're given that the trace is 15 and determinant is 50, so the characteristic polynomial for the matrix in question is

[tex]\lambda^2-15\lambda+50[/tex]

and the eigenvalues are those [tex]\lambda[/tex] for which the characteristic polynomial evaluates to 0.

[tex]\lambda^2-15\lambda+50=(\lambda-5)(\lambda-10)=0\implies\lambda=5,\lambda=10[/tex]
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The eigenvalues of [tex]A[/tex] are 5 and 10.

The eigenvalues of the 2 x 2 matrix are described by the following characteristic polynomial:

[tex]t^{2} - \mathrm{tr} (A)\cdot t +\det(A) = 0[/tex] (1)

The roots of the characteristic polynomial are defined by the quadratic formula:

[tex]t = \frac{\mathrm{tr}(A)\pm \sqrt{[\mathrm{tr}(A)]^{2}-4\cdot \det(A)}}{2}[/tex] (2)

If we know that [tex]\mathrm{tr}(A) = 15[/tex] and [tex]\det(A)= 50[/tex], then the eigenvalues of A are:

[tex]t = \frac{15\pm \sqrt{15^{2}-4\cdot (50)}}{2}[/tex]

[tex]t_{1} = 10\,\lor \,t_{2} = 5[/tex]

The eigenvalues of [tex]A[/tex] are 5 and 10.

We kindly invite to check this question on eigenvalues: https://brainly.com/question/16945116

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