Determine se as seguintes matrizes são inversíveis. Em caso afirmativo, calcule a inversa.
\left(\begin{array}{cccc}1 & 3 & 0 & 0 \\\\ -2 & 1 & 4 & 0 \\\\ 0 & 1 & 2 & 1 \\\\ 0 & 0 & -2 & 5\end{array}\right)

Passo 1

Temos a matriz
A=\begin{pmatrix} 1 & 3 & 0 & 0\\\\ -2 & 1 & 4 & 0\\\\ 0 & 1 & 2 & 1\\\\ 0 & 0 & -2 & 5 \end{pmatrix}
Queremos verificar se \det(A)\neq0 (logo, inversível) e, em caso afirmativo, calcular A^{-1}. Para isso, vamos usar o método de Gauss–Jordan no sistema ampliado (A\mid I_4).

Passo 2

Eliminamos o elemento a_{21}=-2 usando
\mathrm{L}_2\leftarrow \mathrm{L}_2+2\,\mathrm{L}_1.
Na parte ampliada isso faz
\mathrm{L}_2=( -2,1,4,0\mid 0,1,0,0)+2\,(1,3,0,0\mid 1,0,0,0) =(0,7,4,0\mid 2,1,0,0).

Passo 3

Usamos a nova \mathrm{L}_2 para zerar as entradas na coluna 2 em \mathrm{L}_1 e \mathrm{L}_3. Primeiro normalizamos
\mathrm{L}_2\leftarrow \tfrac{1}{7}\,\mathrm{L}_2=(0,1,\tfrac{4}{7},0\mid \tfrac{2}{7},\tfrac{1}{7},0,0).
Depois
\mathrm{L}_1\leftarrow \mathrm{L}_1-3\,\mathrm{L}_2, \quad \mathrm{L}_3\leftarrow \mathrm{L}_3-1\,\mathrm{L}_2,
ficando
\mathrm{L}_1=(1,0,-\tfrac{12}{7},0\mid \tfrac{1}{7},-\tfrac{3}{7},0,0),
\mathrm{L}_3=(0,0,\tfrac{10}{7},1\mid -\tfrac{2}{7},-\tfrac{1}{7},1,0).

Passo 4

Eliminamos a coluna 3. Primeiro normalizamos
\mathrm{L}_3\leftarrow \tfrac{7}{10}\,\mathrm{L}_3 =(0,0,1,\tfrac{7}{10}\mid -\tfrac{1}{5},-\tfrac{1}{10},\tfrac{7}{10},0),
e então
\mathrm{L}_1\leftarrow \mathrm{L}_1+\tfrac{12}{7}\,\mathrm{L}_3, \quad \mathrm{L}_2\leftarrow \mathrm{L}_2-\tfrac{4}{7}\,\mathrm{L}_3, \quad \mathrm{L}_4\leftarrow \mathrm{L}_4+2\,\mathrm{L}_3.
Isso leva a
\mathrm{L}_1=(1,0,0,\tfrac{6}{5}\mid -\tfrac{1}{5},-\tfrac{3}{5},\tfrac{6}{5},0),
\mathrm{L}_2=(0,1,0,-\tfrac{2}{5}\mid \tfrac{2}{5},\tfrac{1}{5},-\tfrac{2}{5},0),
\mathrm{L}_4=(0,0,0,\tfrac{32}{5}\mid -\tfrac{2}{5},-\tfrac{1}{5},\tfrac{7}{5},1).

Passo 5

Eliminamos a coluna 4. Primeiro normalizamos
\mathrm{L}_4\leftarrow \tfrac{5}{32}\,\mathrm{L}_4 =(0,0,0,1\mid -\tfrac{1}{16},-\tfrac{1}{32},\tfrac{7}{32},\tfrac{5}{32}).
Em seguida zeramos acima de a_{44} por
\mathrm{L}_i\leftarrow \mathrm{L}_i - (\text{entrada em coluna 4 de }\mathrm{L}_i)\,\mathrm{L}_4, \quad i=1,2,3.
Após isso todas as colunas à esquerda ficam a matriz identidade e as linhas à direita formam A^{-1}.

Resposta

A^{-1} =\begin{pmatrix} -\tfrac{1}{8} & -\tfrac{9}{16} & \tfrac{15}{16} & -\tfrac{3}{16}\\[6pt] \tfrac{3}{8} & \tfrac{3}{16} & -\tfrac{5}{16} & \tfrac{1}{16}\\[6pt] -\tfrac{5}{32} & -\tfrac{5}{64} & \tfrac{35}{64} & -\tfrac{7}{64}\\[6pt] -\tfrac{1}{16} & -\tfrac{1}{32} & \tfrac{7}{32} & \tfrac{5}{32} \end{pmatrix},
logo \det(A)\neq0 e A é inversível com a inversa acima.]

Exercícios relacionados