Encontre a inversa de cada matriz:
$\left[\begin{array}{cccc}1 & 2 & -3 & 1 \\\\ 0 & -1 & 6 & 2 \\\\ -4 & 0 & 2 & 3 \\\\ 1 & 2 & 3 & 4\end{array}\right]$

O enunciado pede a inversa da matriz

A = \begin{pmatrix} 1 & 2 & -3 & 1\\\\ 0 & -1 & 6 & 2\\\\ -4 & 0 & 2 & 3\\\\ 1 & 2 & 3 & 4 \end{pmatrix}

O procedimento é usar eliminação de Gauss–Jordan: montamos

[A\mid I]

e aplicamos operações elementares até obter

[I\mid A^{-1}]

Formamos o sistema aumentado e zeramos a coluna 1 abaixo de

a_{11}=1

usando

R_3 \leftarrow R_3 + 4\,R_1,\quad R_4 \leftarrow R_4 - R_1.

O resultado é

\left[\begin{array}{cccc|cccc} 1 & 2 & -3 & 1 & 1 & 0 & 0 & 0\\\\ 0 & -1 & 6 & 2 & 0 & 1 & 0 & 0\\\\ 0 & 8 & -10 & 7 & 4 & 0 & 1 & 0\\\\ 0 & 0 & 6 & 3 & -1 & 0 & 0 & 1 \end{array}\right].

Agora o pivô em

(2,2)

-1

Normalizamos: $R_2 \leftarrow -\,R_2\quad\Longrightarrow\quad R_2=(0,1,-6,-2\mid0,-1,0,0).$
Eliminamos na coluna 2: $R_1\leftarrow R_1 - 2\,R_2,\quad R_3\leftarrow R_3 - 8\,R_2.$ Fica $\left[\begin{array}{cccc|cccc} 1 & 0 & 9 & 5 & 1 & 2 & 0 & 0\\\\ 0 & 1 & -6 & -2 & 0 & -1 & 0 & 0\\\\ 0 & 0 & 38 & 23 & 4 & 8 & 1 & 0\\\\ 0 & 0 & 6 & 3 & -1 & 0 & 0 & 1 \end{array}\right].$

O pivô em

(3,3)

vale

38

Normalizamos: $R_3\leftarrow \tfrac1{38}R_3 =(0,0,1,\tfrac{23}{38}\mid\tfrac{2}{19},\tfrac{4}{19},\tfrac{1}{38},0).$
Zeramos a coluna 3 em $R_1,R_2,R_4$ : $R_1\leftarrow R_1 - 9\,R_3,\quad R_2\leftarrow R_2 +6\,R_3,\quad R_4\leftarrow R_4 -6\,R_3.$ O sistema fica $\left[\begin{array}{cccc|cccc} 1 & 0 & 0 & -\tfrac{17}{38} & \tfrac{1}{19} & \tfrac{2}{19} & -\tfrac{9}{38} & 0\\\\ 0 & 1 & 0 & \tfrac{31}{19} & \tfrac{12}{19} & \tfrac{5}{19} & \tfrac{3}{19} & 0\\\\ 0 & 0 & 1 & \tfrac{23}{38} & \tfrac{2}{19} & \tfrac{4}{19} & \tfrac{1}{38} & 0\\\\ 0 & 0 & 0 & -\tfrac{12}{19} & -\tfrac{31}{19} & -\tfrac{24}{19} & -\tfrac{3}{19} & 1 \end{array}\right].$

O pivô em

(4,4)

-\tfrac{12}{19}

Normalizamos: $R_4\leftarrow -\tfrac{19}{12}\,R_4 =(0,0,0,1\mid\tfrac{31}{12},2,\tfrac14,-\tfrac{19}{12}).$
Eliminamos col 4 em $R_1,R_2,R_3$ : $R_1\leftarrow R_1 + \tfrac{17}{38}R_4,\quad R_2\leftarrow R_2 - \tfrac{31}{19}R_4,\quad R_3\leftarrow R_3 - \tfrac{23}{38}R_4.$ O lado esquerdo vira a identidade e o direito é $A^{-1}$ .

A^{-1} = \begin{pmatrix} \tfrac{29}{24} & 1 & -\tfrac18 & -\tfrac{17}{24}\\\\ -\tfrac{43}{12} & -3 & -\tfrac14 & \tfrac{31}{12}\\\\ -\tfrac{35}{24} & -1 & -\tfrac18 & \tfrac{23}{24}\\\ \tfrac{31}{12} & 2 & \tfrac14 & -\tfrac{19}{12} \end{pmatrix}.

Encontre a inversa de cada matriz: \left[\begin{array}{cccc}1 & 2 & -3 & 1 \\\\ 0 & -1 & 6 & 2 \\\\ -4 & 0 & 2 & 3 \\\\ 1 & 2 & 3 & 4\end{array}\right]