![\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]](/tpl/images/0386/9122/c4518.png)
![=det\Bigg(\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\Bigg)\\\\= det\Bigg(\displaystyle\left[\begin{array}{ccc}-14-\lambda&-6&6\\28&12-\lambda&-4\\0&0&4-\lambda\end{array}\right]\Bigg)\\\\=(-14-\lambda)[(12-\lambda)(4-\lambda)]+6[28(4-\lambda)]-6[(28)(0)-(12-\lambda)(0)]\\\\= -\lambda^3 + 2\lambda^2 + 8\lambda](/tpl/images/0386/9122/0403e.png)
![p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda = 0\\-\lambda(\lambda^2-2\lambda-8) = 0\\-\lambda(\lambda^2-4\lambda+2\lambda-8) = 0\\-\lambda[(\lambda(\lambda-4)+2(\lambda-4)] = 0\\-\lambda(\lambda+2)(\lambda-4) = 0 \\\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4](/tpl/images/0386/9122/6b20a.png)
The matrix a = −14 −6 6 28 12 −4 0 0 4 has characteristic polynomial
a. p(λ) = −λ3 + 2 correc...
Mathematics, 22.11.2019 22:31 leahstubbs
The matrix a = −14 −6 6 28 12 −4 0 0 4 has characteristic polynomial
a. p(λ) = −λ3 + 2 correct: your answer is correct.
b. λ2 + incorrect: your answer is incorrect.
c. λ + and therefore the eigenvalues of a are: (arrange the eigenvalues so that λ1 < λ2 < λ3 , and note that all entries here are integers) λ1 = , λ2 = 0 correct: your answer is correct. , λ3 = (hint: one of the roots of p(λ) is -2.)
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