Determine, se existir, a inversa de cada matriz pelo método de Gauss-Jordan:
\mathrm{A}=\left[\begin{array}{ll}5 & 6 \\\\ 4 & 5\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}2 & 5 \\\\ 1 & 3\end{array}\right], C=\left[\begin{array}{ll}1 & 0 \\\\ 0 & 2\end{array}\right] \quad e \mathrm{D}=\left[\begin{array}{cc}1 & -1 \\\\ 1 & 1\end{array}\right]

Passo 1

O enunciado pede para encontrar, quando existir, a inversa de cada matriz 2\times2 pelo método de Gauss–Jordan. Vamos, para cada matriz M, montar a matriz aumentada \bigl[M\mid I\bigr], executar operações elementares até obter \bigl[I\mid M^{-1}\bigr] e, então, ler M^{-1}.

Passo 2

Para A=\begin{pmatrix}5 & 6\\\ 4 & 5\end{pmatrix}, montamos
\left[\begin{array}{cc|cc} 5 & 6 & 1 & 0\\\\ 4 & 5 & 0 & 1 \end{array}\right].
Escalonamos em dois blocos: primeiro obtemos pivô 1 e zeramos abaixo, depois normalizamos e zeramos acima:
R_1\leftarrow \tfrac15R_1,\quad R_2\leftarrow R_2-4R_1 \;\Longrightarrow\; \left[\begin{array}{cc|cc} 1 & \tfrac65 & \tfrac15 & 0\\\\ 0 & \tfrac15 & -\tfrac45 & 1 \end{array}\right],
R_2\leftarrow 5R_2,\quad R_1\leftarrow R_1-\tfrac65\,R_2 \;\Longrightarrow\; \left[\begin{array}{cc|cc} 1 & 0 & 5 & -6\\\\ 0 & 1 & -4 & 5 \end{array}\right].
Logo,
A^{-1}=\begin{pmatrix}5 & -6\\\ -4 & 5\end{pmatrix}.

Passo 3

Para B=\begin{pmatrix}2 & 5\\\ 1 & 3\end{pmatrix}, montamos
\left[\begin{array}{cc|cc} 2 & 5 & 1 & 0\\\\ 1 & 3 & 0 & 1 \end{array}\right].
Escalonando:
R_1\leftarrow \tfrac12R_1,\quad R_2\leftarrow R_2-R_1 \;\Longrightarrow\; \left[\begin{array}{cc|cc} 1 & \tfrac52 & \tfrac12 & 0\\\\ 0 & \tfrac12 & -\tfrac12 & 1 \end{array}\right],
R_2\leftarrow 2R_2,\quad R_1\leftarrow R_1-\tfrac52\,R_2 \;\Longrightarrow\; \left[\begin{array}{cc|cc} 1 & 0 & 3 & -5\\\\ 0 & 1 & -1 & 2 \end{array}\right].
Portanto,
B^{-1}=\begin{pmatrix}3 & -5\\\ -1 & 2\end{pmatrix}.

Passo 4

Para C=\begin{pmatrix}1 & 0\\\ 0 & 2\end{pmatrix}, montamos
\left[\begin{array}{cc|cc} 1 & 0 & 1 & 0\\\\ 0 & 2 & 0 & 1 \end{array}\right].
Aqui já há pivô 1 em R_1. Basta normalizar R_2:
R_2\leftarrow \tfrac12R_2 \;\Longrightarrow\; \left[\begin{array}{cc|cc} 1 & 0 & 1 & 0\\\\ 0 & 1 & 0 & \tfrac12 \end{array}\right].
Como não há nada a zerar em cima de R_2, lemos
C^{-1}=\begin{pmatrix}1 & 0\\\ 0 & \tfrac12\end{pmatrix}.

Passo 5

Para D=\begin{pmatrix}1 & -1\\\ 1 & 1\end{pmatrix}, montamos
\left[\begin{array}{cc|cc} 1 & -1 & 1 & 0\\\\ 1 & 1 & 0 & 1 \end{array}\right].
Escalonando:
R_2\leftarrow R_2-R_1 \;\Longrightarrow\; \left[\begin{array}{cc|cc} 1 & -1 & 1 & 0\\\\ 0 & 2 & -1 & 1 \end{array}\right],
R_2\leftarrow \tfrac12R_2,\quad R_1\leftarrow R_1+R_2 \;\Longrightarrow\; \left[\begin{array}{cc|cc} 1 & 0 & \tfrac12 & \tfrac12\\\\ 0 & 1 & -\tfrac12 & \tfrac12 \end{array}\right].
Assim,
D^{-1}=\begin{pmatrix}\tfrac12 & \tfrac12\\\ -\tfrac12 & \tfrac12\end{pmatrix}.

Resposta

A^{-1}=\begin{pmatrix}5 & -6\\\ -4 & 5\end{pmatrix} B^{-1}=\begin{pmatrix}3 & -5\\\ -1 & 2\end{pmatrix} C^{-1}=\begin{pmatrix}1 & 0\\\ 0 & \tfrac12\end{pmatrix} D^{-1}=\begin{pmatrix}\tfrac12 & \tfrac12\\\ -\tfrac12 & \tfrac12\end{pmatrix}

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