Encontre a inversa se existir
\left[\begin{array}{lll}1 & 2 & 0 \\\\ 2 & 1 & 2 \\\\ 0 & 2 & 1\end{array}\right]

Passo 1

O enunciado pede a inversa da matriz
A = \begin{pmatrix} 1 & 2 & 0\\\\ 2 & 1 & 2\\\\ 0 & 2 & 1 \end{pmatrix}.
O conteúdo é inversão de matriz de ordem 3 usando determinante e matriz adjunta (cofatores e transposta).

Passo 2

Calcular o determinante de A por expansão na primeira linha:
\det(A) =1\cdot\bigl(1\cdot1-2\cdot2\bigr) -2\cdot\bigl(2\cdot1-2\cdot0\bigr) +0\cdot(\cdots) =1\cdot(-3)-2\cdot2 =-3-4 =-7.

Passo 3

Calcular cada menor M_{ij} e o cofator C_{ij}=(-1)^{i+j}M_{ij}: M_{11}=1\cdot1-2\cdot2=-3, logo C_{11}=(-1)^{2}(-3)=-3; M_{12}=2\cdot1-2\cdot0=2, logo C_{12}=(-1)^{3}(2)=-2; M_{13}=2\cdot2-1\cdot0=4, logo C_{13}=(-1)^{4}(4)=4; M_{21}=2\cdot1-0\cdot2=2, logo C_{21}=(-1)^{3}(2)=-2; M_{22}=1\cdot1-0\cdot0=1, logo C_{22}=(-1)^{4}(1)=1; M_{23}=1\cdot2-0\cdot2=2, logo C_{23}=(-1)^{5}(2)=-2; M_{31}=2\cdot2-0\cdot1=4, logo C_{31}=(-1)^{4}(4)=4; M_{32}=1\cdot2-0\cdot2=2, logo C_{32}=(-1)^{5}(2)=-2; M_{33}=1\cdot1-2\cdot2=-3, logo C_{33}=(-1)^{6}(-3)=-3.

Passo 4

Montar a matriz de cofatores e obter a adjunta como sua transposta:
\mathrm{Cof}(A)= \begin{pmatrix} -3 & -2 & 4\\\\ -2 & 1 & -2\\\\ 4 & -2 & -3 \end{pmatrix}, \quad \operatorname{adj}(A)=\bigl(\mathrm{Cof}(A)\bigr)^{T}= \begin{pmatrix} -3 & -2 & 4\\\\ -2 & 1 & -2\\\\ 4 & -2 & -3 \end{pmatrix}.

Passo 5

Como \det(A)=-7, a inversa é
A^{-1}=\frac{1}{\det(A)}\,\operatorname{adj}(A) =-\frac{1}{7} \begin{pmatrix} -3 & -2 & 4\\\\ -2 & 1 & -2\\\\ 4 & -2 & -3 \end{pmatrix} =\frac{1}{7} \begin{pmatrix} 3 & 2 & -4\\\\ 2 & -1 & 2\\\\ -4 & 2 & 3 \end{pmatrix}.

Resposta

A^{-1}=\frac{1}{7} \begin{pmatrix} 3 & 2 & -4\\\\ 2 & -1 & 2\\\\ -4 & 2 & 3 \end{pmatrix}.

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