9.38 Sejam A=\left[\begin{array}{rr}2 & -3 \\\\ 5 & 1\end{array}\right] e B=\left[\begin{array}{ll}1 & 2 \\\\ 0 & 3\end{array}\right]. Encontre f(A), g(A), f(B), g(B), com f(t)=2 t^{2}-5 t+6 e g(t)= t^{3}-2 t^{2}+t+3.

Passo 1

O enunciado pede para avaliar os polinômios matriciais
f(t)=2t^2-5t+6 \quad\text{e}\quad g(t)=t^3-2t^2+t+3
nas matrizes
A=\begin{pmatrix}2 & -3 \\[6pt] 5 & 1\end{pmatrix}, \quad B=\begin{pmatrix}1 & 2 \\[4pt] 0 & 3\end{pmatrix}.
Para isso vamos calcular as potências necessárias de A e B, usar a matriz identidade I, e formar
f(M)=2M^2-5M+6I, \qquad g(M)=M^3-2M^2+M+3I.]

Passo 2

Calculamos as potências de A. Primeiro
A^2 =\begin{pmatrix}2 & -3 \\[4pt] 5 & 1\end{pmatrix} \begin{pmatrix}2 & -3 \\[4pt] 5 & 1\end{pmatrix} =\begin{pmatrix}-11 & -9 \\[4pt] 15 & -14\end{pmatrix}.
Em seguida
A^3 =A\cdot A^2 =\begin{pmatrix}2 & -3 \\[4pt] 5 & 1\end{pmatrix} \begin{pmatrix}-11 & -9 \\[4pt] 15 & -14\end{pmatrix} =\begin{pmatrix}-67 & 24 \\[4pt] -40 & -59\end{pmatrix}.]

Passo 3

Agora formamos
f(A)=2A^2-5A+6I.
Temos
2A^2=\begin{pmatrix}-22 & -18 \\[4pt] 30 & -28\end{pmatrix}, \quad -5A=\begin{pmatrix}-10 & 15 \\[4pt] -25 & -5\end{pmatrix}, \quad 6I=\begin{pmatrix}6 & 0 \\[4pt] 0 & 6\end{pmatrix}.
Somando:
f(A) =\begin{pmatrix}-22 & -18 \\[4pt] 30 & -28\end{pmatrix} +\begin{pmatrix}-10 & 15 \\[4pt] -25 & -5\end{pmatrix} +\begin{pmatrix}6 & 0 \\[4pt] 0 & 6\end{pmatrix} =\begin{pmatrix}-26 & -3 \\[4pt] 5 & -27\end{pmatrix}.]

Passo 4

Para
g(A)=A^3-2A^2+A+3I
usamos os já calculados A^3 e A^2, mais A e 3I:
-2A^2=\begin{pmatrix}22 & 18 \\[4pt] -30 & 28\end{pmatrix}, \quad A=\begin{pmatrix}2 & -3 \\[4pt] 5 & 1\end{pmatrix}, \quad 3I=\begin{pmatrix}3 & 0 \\[4pt] 0 & 3\end{pmatrix}.
Somando em ordem:
A^3-2A^2 =\begin{pmatrix}-45 & 42 \\[4pt] -70 & -31\end{pmatrix}, \quad +\,A =\begin{pmatrix}-43 & 39 \\[4pt] -65 & -30\end{pmatrix}, \quad +\,3I =\begin{pmatrix}-40 & 39 \\[4pt] -65 & -27\end{pmatrix}.
Logo
g(A)=\begin{pmatrix}-40 & 39 \\[4pt] -65 & -27\end{pmatrix}.]

Passo 5

Agora as potências de B. Primeiro
B^2 =\begin{pmatrix}1 & 2 \\[4pt] 0 & 3\end{pmatrix} \begin{pmatrix}1 & 2 \\[4pt] 0 & 3\end{pmatrix} =\begin{pmatrix}1 & 8 \\[4pt] 0 & 9\end{pmatrix},
depois
B^3 =B\cdot B^2 =\begin{pmatrix}1 & 2 \\[4pt] 0 & 3\end{pmatrix} \begin{pmatrix}1 & 8 \\[4pt] 0 & 9\end{pmatrix} =\begin{pmatrix}1 & 26 \\[4pt] 0 & 27\end{pmatrix}.]

Passo 6

Formamos
f(B)=2B^2-5B+6I.
Calculamos
2B^2=\begin{pmatrix}2 & 16 \\[4pt] 0 & 18\end{pmatrix}, \quad -5B=\begin{pmatrix}-5 & -10 \\[4pt] 0 & -15\end{pmatrix}, \quad 6I=\begin{pmatrix}6 & 0 \\[4pt] 0 & 6\end{pmatrix}.
Somando:
f(B) =\begin{pmatrix}2 & 16 \\[4pt] 0 & 18\end{pmatrix} +\begin{pmatrix}-5 & -10 \\[4pt] 0 & -15\end{pmatrix} +\begin{pmatrix}6 & 0 \\[4pt] 0 & 6\end{pmatrix} =\begin{pmatrix}3 & 6 \\[4pt] 0 & 9\end{pmatrix}.]

Passo 7

Finalmente
g(B)=B^3-2B^2+B+3I.
Temos
-2B^2=\begin{pmatrix}-2 & -16 \\[4pt] 0 & -18\end{pmatrix}, \quad B=\begin{pmatrix}1 & 2 \\[4pt] 0 & 3\end{pmatrix}, \quad 3I=\begin{pmatrix}3 & 0 \\[4pt] 0 & 3\end{pmatrix}.
Somando:
B^3-2B^2 =\begin{pmatrix}-1 & 10 \\[4pt] 0 & 9\end{pmatrix}, \quad +\,B =\begin{pmatrix}0 & 12 \\[4pt] 0 & 12\end{pmatrix}, \quad +\,3I =\begin{pmatrix}3 & 12 \\[4pt] 0 & 15\end{pmatrix}.
Portanto
g(B)=\begin{pmatrix}3 & 12 \\[4pt] 0 & 15\end{pmatrix}.]

Resposta

f(A)= \displaystyle\begin{pmatrix}-26 & -3 \\[4pt] 5 & -27\end{pmatrix}
g(A)= \displaystyle\begin{pmatrix}-40 & 39 \\[4pt] -65 & -27\end{pmatrix}
f(B)= \displaystyle\begin{pmatrix}3 & 6 \\[4pt] 0 & 9\end{pmatrix}
g(B)= \displaystyle\begin{pmatrix}3 & 12 \\[4pt] 0 & 15\end{pmatrix}]

Exercícios relacionados