Dadas as matrizes A=\left(\begin{array}{ccc}2 & 1 & 0 \\\\ 0 & 3 & -1 \\\\ 1 & 1 & 4\end{array}\right), B=\left(\begin{array}{lll}1 & 1 & 0 \\\\ 0 & 2 & 2 \\\\ 1 & 0 & 2\end{array}\right) e C=\left(\begin{array}{cc}3 & 5 \\\\ 7 & 10\end{array}\right), calcule: (a) A^{-1}, B^{-1}, C^{-1}. (b) \left[(A-I)^{-1} B\right]^{-1}. (c) 3(A+B) A^{-1}. Passo 1O enunciado fornece as matrizes A = \begin{pmatrix}2 & 1 & 0\\[6pt]0 & 3 & -1\\[6pt]1 & 1 & 4\end{pmatrix},\quad B = \begin{pmatrix}1 & 1 & 0\\[6pt]0 & 2 & 2\\[6pt]1 & 0 & 2\end{pmatrix},\quad C = \begin{pmatrix}3 & 5\\[4pt]7 & 10\end{pmatrix}, e pede em (a) os inversos A^{-1},B^{-1},C^{-1}, em (b) \bigl[(A-I)^{-1}B\bigr]^{-1} e em (c) 3(A+B)A^{-1}. Vamos usar a definição de matriz inversa via adjunta e determinante, a propriedade (XY)^{-1}=Y^{-1}X^{-1} e multiplicação matricial.] Passo 2Calcular A^{-1},B^{-1},C^{-1}. Determinantes: \det A = 25,\quad \det B = 6,\quad \det C = -5. Adjuntas e inversos: A^{-1}=\frac{1}{25} \begin{pmatrix} 13 & -4 & -1\\\\ -1 & 8 & 2\\\\ -3 & -1 & 6 \end{pmatrix}, \quad B^{-1}=\frac{1}{6} \begin{pmatrix} 4 & -2 & 2\\\\ 2 & 2 & -2\\\\ -2 & 1 & 2 \end{pmatrix}, \quad C^{-1}=\frac{1}{-5} \begin{pmatrix} 10 & -5\\\\ -7 & 3 \end{pmatrix} = \begin{pmatrix} -2 & 1\\[4pt] \frac{7}{5} & -\frac{3}{5} \end{pmatrix}.] Passo 3Para \bigl[(A-I)^{-1}B\bigr]^{-1} usamos \bigl[(A-I)^{-1}B\bigr]^{-1} = B^{-1}\,\bigl((A-I)^{-1}\bigr)^{-1} = B^{-1}\,(A-I). Aqui A-I=\begin{pmatrix}1&1&0\\[4pt]0&2&-1\\[4pt]1&1&3\end{pmatrix},\quad B^{-1}=\frac{1}{6}\begin{pmatrix}4&-2&2\\\ 2&2&-2\\\ -2&1&2\end{pmatrix}. Calculando o produto e simplificando: B^{-1}(A-I) =\frac{1}{6} \begin{pmatrix} 6 & 2 & 8\\\\ 0 & 4 & -8\\\\ 0 & 2 & 5 \end{pmatrix} = \begin{pmatrix} 1 & \tfrac{1}{3} & \tfrac{4}{3}\\[4pt] 0 & \tfrac{2}{3} & -\tfrac{4}{3}\\[4pt] 0 & \tfrac{1}{3} & \tfrac{5}{6} \end{pmatrix}.] Passo 4Para 3(A+B)A^{-1} primeiro A+B =\begin{pmatrix}3&2&0\\[4pt]0&5&1\\[4pt]2&1&6\end{pmatrix}, \quad A^{-1}=\frac{1}{25} \begin{pmatrix} 13 & -4 & -1\\\\ -1 & 8 & 2\\\\ -3 & -1 & 6 \end{pmatrix}. Multiplicando e depois multiplicando por 3: (A+B)A^{-1} =\frac{1}{25} \begin{pmatrix} 37 & 4 & 1\\\\ -8 & 39 & 16\\\\ 7 & -6 & 36 \end{pmatrix}, 3(A+B)A^{-1} =\frac{3}{25} \begin{pmatrix} 37 & 4 & 1\\\\ -8 & 39 & 16\\\\ 7 & -6 & 36 \end{pmatrix}.] Resposta(a) A^{-1}=\frac{1}{25} \begin{pmatrix} 13 & -4 & -1\\\\ -1 & 8 & 2\\\\ -3 & -1 & 6 \end{pmatrix},\quad B^{-1}=\frac{1}{6} \begin{pmatrix} 4 & -2 & 2\\\\ 2 & 2 & -2\\\\ -2 & 1 & 2 \end{pmatrix},\quad C^{-1}= \begin{pmatrix} -2 & 1\\[4pt] \frac{7}{5} & -\frac{3}{5} \end{pmatrix}. (b) \bigl[(A-I)^{-1}B\bigr]^{-1} = \begin{pmatrix} 1 & \tfrac{1}{3} & \tfrac{4}{3}\\[4pt] 0 & \tfrac{2}{3} & -\tfrac{4}{3}\\[4pt] 0 & \tfrac{1}{3} & \tfrac{5}{6} \end{pmatrix}. (c) 3(A+B)A^{-1} =\frac{3}{25} \begin{pmatrix} 37 & 4 & 1\\\\ -8 & 39 & 16\\\\ 7 & -6 & 36 \end{pmatrix}.] MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosDadas as matrizes A=\left(\begin{array}{ccc}2 & 1 & 0 \\\\ 0 & 3 & -1 \\\\ 1 & 1 & 4\end{array}\right), B=\left(\begin{array}{lll}1 & 1 & 0 \\\\ 0 & 2 & 2 \\\\ 1 & 0 & 2\end{array}\right) e C=\left(\begin{array}{cc}3 & 5 \\\\ 7 & 10\end{array}\right), calcule: (a) A^{-1}, B^{-1}, C^{-1}. (b) \left[(A-I)^{-1} B\right]^{-1}. (c) 3(A+B) A^{-1}. Dadas as matrizes A=\left(\begin{array}{ccc}2 & 1 & 0 \\\\ 0 & 3 & -1 \\\\ 1 & 1 & 4\end{array}\right), B=\left(\begin{array}{lll}1 & 1 & 0 \\\\ 0 & 2 & 2 \\\\ 1 & 0 & 2\end{array}\right) e C=\left(\begin{array}{cc}3 & 5 \\\\ 7 & 10\end{array}\right), calcule: (a) A^{-1}, B^{-1}, C^{-1}. (b) \left[(A-I)^{-1} B\right]^{-1}. (c) 3(A+B) A^{-1}. (a) Seja P=(a, b). Determine o lugar geométrico dos pontos do plano que distam r de P. (b) Seja P=(a, b, c). Determine o lugar geométrico dos pontos do espaço que distam r do ponto P. Resolução de 18: Chamemos de \vec{u}=(1,1,3) e \vec{v}=(0,1,-1) os vetores diretores das retas r e s, respectivamente. Se a reta m é ortogonal a r e s podemos considerar \vec{n}=\vec{u} \times \vec{v} com seu vetor diretor. Então, \vec{n}=\vec{u} \times \vec{v} \equiv \operatorname{det}\left[\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\\\ 1 & 1 & 3 \\\\ 0 & 1 & -1 \end{array}\right] \equiv(-4,1,1) Como A(4,5,1) \in m, podemos escrever m:(x, y, z)=(4,5,1)+\mu(-4,1,1), \mu \in R, como uma equação vetorial de m .