Considere as seguintes matrizes:
A=\left[\begin{array}{lll}1 & 2 & 1 \\\\ 0 & 1 & 2 \\\\ 1 & 1 & 1\end{array}\right] \quad B=\left[\begin{array}{ccc}1 & 1 & -1 \\\\ 2 & 1 & 1 \\\\ 3 & -1 & 1\end{array}\right]
A B é inversível? Se sim, determine (A B)^{-1}. (Use o conceito de operações elementares).

Passo 1

O enunciado dá as matrizes
A = \begin{pmatrix} 1 & 2 & 1\\\\ 0 & 1 & 2\\\\ 1 & 1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 & -1\\\\ 2 & 1 & 1\\\\ 3 & -1 & 1 \end{pmatrix}
e pergunta se AB é invertível e, em caso afirmativo, pede (AB)^{-1}.
Como regra geral, se A e B são invertíveis então AB também é, com
(AB)^{-1} = B^{-1}A^{-1}.
Para mostrar que A e B são invertíveis e achar A^{-1} e B^{-1}, vamos usar operações elementares (método da matriz aumentada).

Passo 2

Calcular A^{-1} montando \bigl[A\mid I_3\bigr] e aplicando eliminações:
\left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0\\\\ 0 & 1 & 2 & 0 & 1 & 0\\\\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{R_3\to R_3-R_1}\; \left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0\\\\ 0 & 1 & 2 & 0 & 1 & 0\\\\ 0 & -1& 0 & -1& 0 & 1 \end{array}\right]
\xrightarrow{R_3\to R_3+R_2}\quad \left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0\\\\ 0 & 1 & 2 & 0 & 1 & 0\\\\ 0 & 0 & 2 & -1& 1 & 1 \end{array}\right] \;\xrightarrow{R_3\to \tfrac12R_3}\; \left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0\\\\ 0 & 1 & 2 & 0 & 1 & 0\\\\ 0 & 0 & 1 & -\tfrac12& \tfrac12& \tfrac12 \end{array}\right]
\xrightarrow{\substack{R_1\to R_1-2\,R_2\R_2\to R_2-2\,R_3\R_1\to R_1+3\,R_3}} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -\tfrac12 & -\tfrac12 & \tfrac32\\\\ 0 & 1 & 0 & 1 & 0 & -1\\\\ 0 & 0 & 1 & -\tfrac12 & \tfrac12 & \tfrac12 \end{array}\right].
Logo
A^{-1} = \begin{pmatrix} -\tfrac12 & -\tfrac12 & \tfrac32\\\\ 1 & 0 & -1\\\\ -\tfrac12 & \tfrac12 & \tfrac12 \end{pmatrix}.

Passo 3

Calcular B^{-1} montando \bigl[B\mid I_3\bigr]:
\left[\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 0 & 0\\\\ 2 & 1 & 1 & 0 & 1 & 0\\\\ 3 & -1 & 1 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{\substack{R_2\to R_2-2\,R_1\R_3\to R_3-3\,R_1}} \left[\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 0 & 0\\\\ 0 & -1 & 3 & -2 & 1 & 0\\\\ 0 & -4 & 4 & -3 & 0 & 1 \end{array}\right]
\xrightarrow{R_2\to -\,R_2}\; \left[\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 0 & 0\\\\ 0 & 1 & -3 & 2 & -1& 0\\\\ 0 & -4& 4 & -3 & 0 & 1 \end{array}\right] \;\xrightarrow{R_3\to R_3+4\,R_2}\; \left[\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 0 & 0\\\\ 0 & 1 & -3 & 2 & -1& 0\\\\ 0 & 0 & -8 & 5 & -4& 1 \end{array}\right]
\xrightarrow{R_3\to -\tfrac18R_3}\; \left[\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 0 & 0\\\\ 0 & 1 & -3 & 2 & -1 & 0\\\\ 0 & 0 & 1 & -\tfrac58& \tfrac12 & -\tfrac18 \end{array}\right]
\xrightarrow{\substack{R_1\to R_1+R_3\R_2\to R_2-(-3)R_3}} \left[\begin{array}{ccc|ccc} 1 & 1 & 0 & \tfrac14 & 0 & \tfrac14\\\\ 0 & 1 & 0 & \tfrac18 & \tfrac12& -\tfrac38\\\\ 0 & 0 & 1 & -\tfrac58& \tfrac12& -\tfrac18 \end{array}\right].
Finalmente
B^{-1} = \begin{pmatrix} \tfrac14 & 0 & \tfrac14\\\ \tfrac18 & \tfrac12 & -\tfrac38\\\\ -\tfrac58 & \tfrac12 & -\tfrac18 \end{pmatrix}.

Passo 4

Como A e B são invertíveis, AB também é. E vale
(AB)^{-1} \;=\; B^{-1}\,A^{-1}.

Passo 5

Multiplicando
B^{-1}A^{-1} =\begin{pmatrix} \tfrac14 & 0 & \tfrac14\\\ \tfrac18 & \tfrac12 & -\tfrac38\\\\ -\tfrac58 & \tfrac12 & -\tfrac18 \end{pmatrix} \begin{pmatrix} -\tfrac12 & -\tfrac12 & \tfrac32\\\\ 1 & 0 & -1\\\\ -\tfrac12 & \tfrac12 & \tfrac12 \end{pmatrix},
obtém-se
(AB)^{-1} =\begin{pmatrix} -\tfrac14 & 0 & \tfrac12\\\ \tfrac58 & -\tfrac14 & -\tfrac12\\\ \tfrac78 & \tfrac14 & -\tfrac32 \end{pmatrix}.

Resposta

Sim, AB é invertível e
(AB)^{-1} =\begin{pmatrix} -\tfrac14 & 0 & \tfrac12\\\ \tfrac58 & -\tfrac14 & -\tfrac12\\\ \tfrac78 & \tfrac14 & -\tfrac32 \end{pmatrix}.

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