3^{\circ} ) Sendo A=\left[\begin{array}{cc}4 & 2 \\\\ 3 & -1\end{array}\right] e I_{2} a matriz identidade de ordem 2.
a) Calcule os valores de \lambda \in R, de modo que \operatorname{det}\left(A-\lambda I_{2}\right)=0
b) Determine a matriz D tal que D=P^{-1} A P, onde a matriz \mathrm{P}=\left[\begin{array}{cc}3-\lambda_{i} & -4 \\\\ 2 & -6-\lambda_{i}\end{array}\right] para - menor dos \lambda_{i} calculado no item ( a ).

Passo 1

O enunciado pede que, para a matriz
A = \begin{pmatrix}4 & 2\\\ 3 & -1\end{pmatrix},
a) encontremos os valores de \lambda\in\mathbb{R} que anulam \det(A-\lambda I_2), ou seja, os autovalores de A;
b) escolhendo o menor desses autovalores \lambda_i, montemos
P=\begin{pmatrix}3-\lambda_i & -4\\\ 2 & -6-\lambda_i\end{pmatrix}
e calculemos a matriz
D = P^{-1} A P.

Passo 2

Para achar os autovalores calculamos o polinômio característico:
\det(A-\lambda I_2) =\det\begin{pmatrix}4-\lambda & 2\\\ 3 & -1-\lambda\end{pmatrix} =(4-\lambda)(-1-\lambda)-2\cdot3.
Expandindo,
(4-\lambda)(-1-\lambda)-6 =-(4-\lambda)(1+\lambda)-6 =-\bigl(4+4\lambda-\lambda-\lambda^2\bigr)-6 =\lambda^2-3\lambda-10.
Portanto resolvemos
\lambda^2-3\lambda-10=0 \quad\Longrightarrow\quad \lambda=\frac{3\pm\sqrt{9+40}}{2} =\frac{3\pm7}{2},
isto é,
\lambda_1=5, \quad \lambda_2=-2.

Passo 3

O menor autovalor é \lambda_i=\lambda_2=-2. Substituímos em
P=\begin{pmatrix}3-\lambda_i & -4\\\ 2 & -6-\lambda_i\end{pmatrix}
para obter
P=\begin{pmatrix}3-(-2) & -4\\\ 2 & -6-(-2)\end{pmatrix} =\begin{pmatrix}5 & -4\\\ 2 & -4\end{pmatrix}.

Passo 4

Calculamos \det P e a inversa P^{-1}.
\det P = 5\cdot(-4)-(-4)\cdot2=-20+8=-12,
P^{-1}=\frac1{\det P}\begin{pmatrix}-4 & 4\\\ -2 & 5\end{pmatrix} =\frac1{-12}\begin{pmatrix}-4 & 4\\\ -2 & 5\end{pmatrix} =\begin{pmatrix}1/3 & -1/3\\\ 1/6 & -5/12\end{pmatrix}.

Passo 5

Multiplicamos A por P:
A\,P =\begin{pmatrix}4 & 2\\\ 3 & -1\end{pmatrix} \begin{pmatrix}5 & -4\\\ 2 & -4\end{pmatrix} =\begin{pmatrix}4\cdot5+2\cdot2 & 4\cdot(-4)+2\cdot(-4)\\\\ 3\cdot5+(-1)\cdot2 & 3\cdot(-4)+(-1)\cdot(-4)\end{pmatrix} =\begin{pmatrix}24 & -24\\\ 13 & -8\end{pmatrix}.

Passo 6

Finalmente,
D = P^{-1}(A\,P) =\begin{pmatrix}1/3 & -1/3\\\ 1/6 & -5/12\end{pmatrix} \begin{pmatrix}24 & -24\\\ 13 & -8\end{pmatrix}.
Calculando as entradas de D:
D_{11}=\tfrac13\cdot24+(-\tfrac13)\cdot13=\frac{24-13}{3}=\frac{11}{3}, \quad D_{12}=\tfrac13\cdot(-24)+(-\tfrac13)\cdot(-8)=\frac{-24+8}{3}=-\frac{16}{3},
D_{21}=\tfrac16\cdot24+(-\tfrac5{12})\cdot13 =4-\frac{65}{12} =-\frac{17}{12}, \quad D_{22}=\tfrac16\cdot(-24)+(-\tfrac5{12})\cdot(-8) =-4+\frac{40}{12} =-\frac{2}{3}.

Resposta

a) \lambda_1=5 e \lambda_2=-2.
b) Para \lambda_i=-2 obtemos
P=\begin{pmatrix}5 & -4\\\ 2 & -4\end{pmatrix}, \quad D=P^{-1}AP =\begin{pmatrix}11/3 & -16/3\\\ -17/12 & -2/3\end{pmatrix}.

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