Dada as matrizes:
A=\left[\begin{array}{lll} 1 & 2 & 1 \\\\ 0 & 1 & 2 \\\\ 1 & 1 & 1 \end{array}\right] \quad B=\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 0 & 1 & 2 \\\\ 1 & 3 & 4 \end{array}\right] \quad C=\left[\begin{array}{rrr} 2 & 0 & -1 \\\\ 3 & 0 & 2 \\\\ 4 & -3 & 7 \end{array}\right] \quad D=\left[\begin{array}{rrrr} 3 & -1 & 5 & 0 \\\\ 0 & 2 & 0 & 1 \\\\ 2 & 0 & -1 & 3 \\\\ 1 & 1 & 2 & 0 \end{array}\right]
Calcule a matrix adjunta para cada matriz.

Passo 1

O que o enunciado pede e o que é dado.
  • Pede: calcular a matriz adjunta (adjugate) de cada uma das matrizes A, B, C e D.
  • Matrizes fornecidas:
A=\left[\begin{array}{lll} 1 & 2 & 1 \\\\ 0 & 1 & 2 \\\\ 1 & 1 & 1 \end{array}\right],\quad B=\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 0 & 1 & 2 \\\\ 1 & 3 & 4 \end{array}\right],
C=\left[\begin{array}{rrr} 2 & 0 & -1 \\\\ 3 & 0 & 2 \\\\ 4 & -3 & 7 \end{array}\right],\quad D=\left[\begin{array}{rrrr} 3 & -1 & 5 & 0 \\\\ 0 & 2 & 0 & 1 \\\\ 2 & 0 & -1 & 3 \\\\ 1 & 1 & 2 & 0 \end{array}\right].

Passo 2

O que é a “matriz adjunta” e como vamos calcular.
  • Definição prática (a que usaremos): a matriz adjunta de uma matriz quadrada M é a transposta da matriz dos cofatores de M. Em notação:
\operatorname{adj}(M)=\big(\,C_{ij}\,\big)^{\!T},
onde
C_{ij}=(-1)^{i+j}\det\big(M_{ij}\big)
e M_{ij} é a matriz que se obtém de M ao eliminar a linha i e a coluna j (chamada menor de ordem n-1).
  • Portanto o procedimento para cada matriz: para cada par (i,j) calcular o menor M_{ij}, calcular seu determinante, multiplicar por (-1)^{i+j} para obter o cofator C_{ij}, montar a matriz de cofatores e, por fim, transpor essa matriz para obter \operatorname{adj}(M). Vou fazer isso passo a passo para A,B,C,D.

Passo 3

Cálculo de \operatorname{adj}(A). Primeiro calculo todos os menores de ordem 2 (eliminação de uma linha e uma coluna) e os cofatores para A.
  • Matriz A:
A=\left[\begin{array}{ccc} 1 & 2 & 1\\\\ 0 & 1 & 2\\\\ 1 & 1 & 1 \end{array}\right].
  • Menores e determinantes (cada \det(M_{ij})):
    • M_{11}=\left[\begin{array}{cc}1 & 2\\[2pt]1 & 1\end{array}\right]\Rightarrow\det(M_{11})=1\cdot1-2\cdot1=-1. Então C_{11}=(-1)^{1+1}(-1)=-1.
    • M_{12}=\left[\begin{array}{cc}0 & 2\\[2pt]1 & 1\end{array}\right]\Rightarrow\det(M_{12})=0\cdot1-2\cdot1=-2. Então C_{12}=(-1)^{1+2}(-2)=2.
    • M_{13}=\left[\begin{array}{cc}0 & 1\\[2pt]1 & 1\end{array}\right]\Rightarrow\det(M_{13})=0\cdot1-1\cdot1=-1. Então C_{13}=(-1)^{1+3}(-1)=-1.
    • M_{21}=\left[\begin{array}{cc}2 & 1\\[2pt]1 & 1\end{array}\right]\Rightarrow\det(M_{21})=2\cdot1-1\cdot1=1. Então C_{21}=(-1)^{2+1}(1)=-1.
    • M_{22}=\left[\begin{array}{cc}1 & 1\\[2pt]1 & 1\end{array}\right]\Rightarrow\det(M_{22})=1\cdot1-1\cdot1=0. Então C_{22}=0.
    • M_{23}=\left[\begin{array}{cc}1 & 2\\[2pt]1 & 1\end{array}\right]\Rightarrow\det(M_{23})=1\cdot1-2\cdot1=-1. Então C_{23}=(-1)^{2+3}(-1)=1.
    • M_{31}=\left[\begin{array}{cc}2 & 1\\[2pt]1 & 2\end{array}\right]\Rightarrow\det(M_{31})=2\cdot2-1\cdot1=3. Então C_{31}=3.
    • M_{32}=\left[\begin{array}{cc}1 & 1\\[2pt]0 & 2\end{array}\right]\Rightarrow\det(M_{32})=1\cdot2-1\cdot0=2. Então C_{32}=(-1)^{3+2}(2)=-2.
    • M_{33}=\left[\begin{array}{cc}1 & 2\\[2pt]0 & 1\end{array}\right]\Rightarrow\det(M_{33})=1\cdot1-2\cdot0=1. Então C_{33}=1.
  • Matriz dos cofatores:
\big(C_{ij}\big)=\left[\begin{array}{rrr} -1 & 2 & -1\\\\ -1 & 0 & 1\\\\ 3 & -2 & 1 \end{array}\right].
  • Transpondo para obter a adjunta:
\operatorname{adj}(A)=\left[\begin{array}{rrr} -1 & -1 & 3\\\\ 2 & 0 & -2\\\\ -1 & 1 & 1 \end{array}\right].]

Passo 4

Cálculo de \operatorname{adj}(B). Mesma rotina para B.
  • Matriz B:
B=\left[\begin{array}{ccc} 1 & 2 & 2\\\\ 0 & 1 & 2\\\\ 1 & 3 & 4 \end{array}\right].
  • Menores, determinantes e cofatores:
    • M_{11}=\left[\begin{array}{cc}1 & 2\\[2pt]3 & 4\end{array}\right]\Rightarrow\det=-2,\;C_{11}=-2.
    • M_{12}=\left[\begin{array}{cc}0 & 2\\[2pt]1 & 4\end{array}\right]\Rightarrow\det=-2,\;C_{12}=2.
    • M_{13}=\left[\begin{array}{cc}0 & 1\\[2pt]1 & 3\end{array}\right]\Rightarrow\det=-1,\;C_{13}=-1.
    • M_{21}=\left[\begin{array}{cc}2 & 2\\[2pt]3 & 4\end{array}\right]\Rightarrow\det=2,\;C_{21}=-2.
    • M_{22}=\left[\begin{array}{cc}1 & 2\\[2pt]1 & 4\end{array}\right]\Rightarrow\det=2,\;C_{22}=2.
    • M_{23}=\left[\begin{array}{cc}1 & 2\\[2pt]1 & 3\end{array}\right]\Rightarrow\det=1,\;C_{23}=-1.
    • M_{31}=\left[\begin{array}{cc}2 & 2\\[2pt]1 & 2\end{array}\right]\Rightarrow\det=2,\;C_{31}=2.
    • M_{32}=\left[\begin{array}{cc}1 & 2\\[2pt]0 & 2\end{array}\right]\Rightarrow\det=2,\;C_{32}=-2.
    • M_{33}=\left[\begin{array}{cc}1 & 2\\[2pt]0 & 1\end{array}\right]\Rightarrow\det=1,\;C_{33}=1.
  • Matriz dos cofatores:
\big(C_{ij}\big)=\left[\begin{array}{rrr} -2 & 2 & -1\\\\ -2 & 2 & -1\\\\ 2 & -2 & 1 \end{array}\right].
  • Transpondo:
\operatorname{adj}(B)=\left[\begin{array}{rrr} -2 & -2 & 2\\\\ 2 & 2 & -2\\\\ -1 & -1 & 1 \end{array}\right].
(Observação: \det(B)=0, então B é singular; a adjunta ainda existe mas não permite formar B^{-1}.)]

Passo 5

Cálculo de \operatorname{adj}\text{(C)}.
  • Matriz C:
C=\left[\begin{array}{rrr} 2 & 0 & -1\\\\ 3 & 0 & 2\\\\ 4 & -3 & 7 \end{array}\right].
  • Menores, determinantes e cofatores:
    • M_{11}=\left[\begin{array}{cc}0 & 2\\[2pt]-3 & 7\end{array}\right]\Rightarrow\det=0\cdot7-2\cdot(-3)=6,\;C_{11}=6.
    • M_{12}=\left[\begin{array}{cc}3 & 2\\[2pt]4 & 7\end{array}\right]\Rightarrow\det=13,\;C_{12}=-13.
    • M_{13}=\left[\begin{array}{cc}3 & 0\\[2pt]4 & -3\end{array}\right]\Rightarrow\det=-9,\;C_{13}=-9.
    • M_{21}=\left[\begin{array}{cc}0 & -1\\[2pt]-3 & 7\end{array}\right]\Rightarrow\det=-3,\;C_{21}=3.
    • M_{22}=\left[\begin{array}{cc}2 & -1\\[2pt]4 & 7\end{array}\right]\Rightarrow\det=18,\;C_{22}=18.
    • M_{23}=\left[\begin{array}{cc}2 & 0\\[2pt]4 & -3\end{array}\right]\Rightarrow\det=-6,\;C_{23}=6.
    • M_{31}=\left[\begin{array}{cc}0 & -1\\[2pt]0 & 2\end{array}\right]\Rightarrow\det=0,\;C_{31}=0.
    • M_{32}=\left[\begin{array}{cc}2 & -1\\[2pt]3 & 2\end{array}\right]\Rightarrow\det=7,\;C_{32}=-7.
    • M_{33}=\left[\begin{array}{cc}2 & 0\\[2pt]3 & 0\end{array}\right]\Rightarrow\det=0,\;C_{33}=0.
  • Matriz dos cofatores:
\big(C_{ij}\big)=\left[\begin{array}{rrr} 6 & -13 & -9\\\\ 3 & 18 & 6\\\\ 0 & -7 & 0 \end{array}\right].
  • Transpondo:
\operatorname{adj}\text{(C)}=\left[\begin{array}{rrr} 6 & 3 & 0\\\\ -13 & 18 & -7\\\\ -9 & 6 & 0 \end{array}\right].]

Passo 6

Cálculo de \operatorname{adj}(D) (passo a passo, 4×4).
  • Matriz D:
D=\left[\begin{array}{rrrr} 3 & -1 & 5 & 0\\\\ 0 & 2 & 0 & 1\\\\ 2 & 0 & -1 & 3\\\\ 1 & 1 & 2 & 0 \end{array}\right].
  • Para cada C_{ij}=(-1)^{i+j}\det(M_{ij}) calculei o menor M_{ij} (matriz 3\times3) e seu determinante. Apresento os cofatores finais (cálculos detalhados feitos termo a termo):
\big(C_{ij}\big)=\left[\begin{array}{rrrr} -11 & -5 & 8 & 10\\\\ -21 & -3 & 12 & 18\\\\ 7 & 1 & -4 & -2\\\\ 31 & 13 & -16 & -26 \end{array}\right].
  • Transpondo para obter a adjunta:
\operatorname{adj}(D)=\left[\begin{array}{rrrr} -11 & -21 & 7 & 31\\\\ -5 & -3 & 1 & 13\\\\ 8 & 12 & -4 & -16\\\\ 10 & 18 & -2 & -26 \end{array}\right].
(Verificação rápida: \det(D)=3\cdot(-11)+(-1)\cdot(-5)+5\cdot8+0\cdot10= -33+5+40=12, consistente com D\,\operatorname{adj}(D)=\det(D)I.)

Resposta

\operatorname{adj}(A)=\left[\begin{array}{rrr} -1 & -1 & 3\\\\ 2 & 0 & -2\\\\ -1 & 1 & 1 \end{array}\right]
\operatorname{adj}(B)=\left[\begin{array}{rrr} -2 & -2 & 2\\\\ 2 & 2 & -2\\\\ -1 & -1 & 1 \end{array}\right]
\operatorname{adj}\text{(C)}=\left[\begin{array}{rrr} 6 & 3 & 0\\\\ -13 & 18 & -7\\\\ -9 & 6 & 0 \end{array}\right]
\operatorname{adj}(D)=\left[\begin{array}{rrrr} -11 & -21 & 7 & 31\\\\ -5 & -3 & 1 & 13\\\\ 8 & 12 & -4 & -16\\\\ 10 & 18 & -2 & -26 \end{array}\right]

Exercícios relacionados