Question

Dada a matriz A = $\left[\begin{array}{ccc}1&5\\2&4\\\end{array}\right]$, calcule os autovalores da matriz A, logo em seguida marque a alternativa correta:




a. λ $_{1}$ =5 e λ $_{2}$= 2


b. λ $_{1}$ =- 1 e λ $_{2}$= 7


c. λ $_{1}$= - 4 e λ $_{2}$= - 1


d. λ $_{1}$= - 1 e λ $_{2}$= 6


e. λ $_{1}$= - 3 e λ $_{2}$= 7

Answer 1

Olá



Alternativa correta, letra D) λ = -1, λ = 6



Podemos calcular os autovalores de uma matriz resolvendo o...:

$\displaystyle \mathsf{det\left(\lambda I-A\right)=0}$



Onde:
'I' é a matriz identidade (de mesma ordem da matriz 'A')
'A' é a matriz


Calculando então


$\displaystyle \mathsf{det\left(\lambda \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] -\left[\begin{array}{ccc}1&5\\2&4\\\end{array}\right]\right)=0}\\\\\\\\\text{Efetutando a multiplicacao entre }\lambda~e~I\\\\\\\\\mathsf{det\left( \left[\begin{array}{ccc}\lambda&0\\0&\lambda\\\end{array}\right] -\left[\begin{array}{ccc}1&5\\2&4\\\end{array}\right]\right)=0}\\\\\\\\\text{Efetuando a subtracao das duas matrizes}$


$\displaystyle \mathsf{det\left( \left[\begin{array}{ccc}\lambda -1&-5\\-2&\lambda-4\\\end{array}\right] \right)=0}$


Calculando o determinante da matriz e igualando à zero.


$\displaystyle \mathsf{det\left( \left[\begin{array}{ccc}\lambda -1&-5\\-2&\lambda-4\\\end{array}\right] \right)=0}\\\\\\\\\mathsf{\underbrace{(\mathsf{(\lambda -1)\cdot(\lambda -4)})}_{diag.~principal}~~-~~\underbrace{(\mathsf{-2})\cdot(-5)}_{diag.~secund\'aria}~=~0}\\\\\\\\\mathsf{\lambda ^2-4\lambda-\lambda +4-10=0}\\\\\\\mathsf{\lambda^2-5\lambda-6=0}$


Agora basta resolver essa equação do 2º grau, com isso chegaremos nos autovalores da matriz A.


$\displaystyle \mathsf{\lambda^2-5\lambda-6=0}\\\\\\\mathsf{\Delta=49}\\\\\\\mathsf{\lambda= \frac{-(-5)\pm \sqrt{49} }{2\cdot 1} }\\\\\\\\\boxed{\mathsf{\lambda_1=-1}}\\\\\\\boxed{\mathsf{\lambda_2=6}}$





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