1. **Problema:** Escrever a matriz $M = (a_{ij})$ de ordem $2 \times 3$ com as regras dadas: - Para $i = j$, $a_{ij} = \frac{1}{2}(i^2 - j) + i$ - Para $i \neq j$, $a_{ij} = (i + j)^2 - 2j$ **Passos:** 1. A matriz $M$ tem 2 linhas ($i=1,2$) e 3 colunas ($j=1,2,3$). 2. Calcular cada elemento: - $a_{11}$: $i=j=1$, então $a_{11} = \frac{1}{2}(1^2 - 1) + 1 = \frac{1}{2}(0) + 1 = 1$ - $a_{12}$: $i=1 \neq j=2$, então $a_{12} = (1+2)^2 - 2\times 2 = 3^2 - 4 = 9 - 4 = 5$ - $a_{13}$: $i=1 \neq j=3$, então $a_{13} = (1+3)^2 - 2\times 3 = 4^2 - 6 = 16 - 6 = 10$ - $a_{21}$: $i=2 \neq j=1$, então $a_{21} = (2+1)^2 - 2\times 1 = 3^2 - 2 = 9 - 2 = 7$ - $a_{22}$: $i=j=2$, então $a_{22} = \frac{1}{2}(2^2 - 2) + 2 = \frac{1}{2}(4 - 2) + 2 = \frac{1}{2}(2) + 2 = 1 + 2 = 3$ - $a_{23}$: $i=2 \neq j=3$, então $a_{23} = (2+3)^2 - 2\times 3 = 5^2 - 6 = 25 - 6 = 19$ 3. Matriz $M$: $$ M = \begin{pmatrix} 1 & 5 & 10 \\ 7 & 3 & 19 \end{pmatrix} $$ --- 2. **Problema:** Dadas as matrizes $$ A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \\ -2 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 2 & 3 \end{pmatrix} $$ Calcular: **a)** $A - 2B^T$ **Passos:** 1. Calcular $B^T$ (transposta de $B$): $$ B^T = \begin{pmatrix} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{pmatrix} $$ 2. Multiplicar $B^T$ por 2: $$ 2B^T = \begin{pmatrix} 2 & 0 \\ -2 & 4 \\ 0 & 6 \end{pmatrix} $$ 3. $A$ é $3 \times 2$ e $2B^T$ é $3 \times 2$, então podemos subtrair: $$ A - 2B^T = \begin{pmatrix} 2-2 & 1-0 \\ 1-(-2) & 3-4 \\ -2-0 & 0-6 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & -1 \\ -2 & -6 \end{pmatrix} $$ **b)** Calcular $(A + B^T) \cdot (A^T - B)$ 1. Calcular $A + B^T$: $$ A + B^T = \begin{pmatrix} 2+1 & 1+0 \\ 1+(-1) & 3+2 \\ -2+0 & 0+3 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 0 & 5 \\ -2 & 3 \end{pmatrix} $$ 2. Calcular $A^T$: $$ A^T = \begin{pmatrix} 2 & 1 & -2 \\ 1 & 3 & 0 \end{pmatrix} $$ 3. Calcular $A^T - B$ (ambos $2 \times 3$): $$ A^T - B = \begin{pmatrix} 2-1 & 1-(-1) & -2-0 \\ 1-0 & 3-2 & 0-3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & -2 \\ 1 & 1 & -3 \end{pmatrix} $$ 4. Multiplicar $(A + B^T)_{3\times 2}$ por $(A^T - B)_{2\times 3}$: $$ \begin{pmatrix} 3 & 1 \\ 0 & 5 \\ -2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & -2 \\ 1 & 1 & -3 \end{pmatrix} = \begin{pmatrix} 3\times1 + 1\times1 & 3\times2 + 1\times1 & 3\times(-2) + 1\times(-3) \\ 0\times1 + 5\times1 & 0\times2 + 5\times1 & 0\times(-2) + 5\times(-3) \\ -2\times1 + 3\times1 & -2\times2 + 3\times1 & -2\times(-2) + 3\times(-3) \end{pmatrix} = \begin{pmatrix} 4 & 7 & -9 \\ 5 & 5 & -15 \\ 1 & -1 & -5 \end{pmatrix} $$ --- 3. **Problema:** Resolver o sistema matricial $$ \begin{cases} x + y = A - 2B \\ x - y = 2A + B \end{cases} $$ com $$ A = \begin{pmatrix} 3 & 1 \\ 3 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} $$ **Passos:** 1. Somar as duas equações para eliminar $y$: $$ (x + y) + (x - y) = (A - 2B) + (2A + B) \Rightarrow 2x = 3A - B $$ 2. Subtrair a segunda da primeira para eliminar $x$: $$ (x + y) - (x - y) = (A - 2B) - (2A + B) \Rightarrow 2y = -A - 3B $$ 3. Calcular $3A - B$: $$ 3A = \begin{pmatrix} 9 & 3 \\ 9 & -3 \end{pmatrix}, \quad 3A - B = \begin{pmatrix} 9-2 & 3-1 \\ 9-3 & -3-0 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 6 & -3 \end{pmatrix} $$ 4. Calcular $-A - 3B$: $$ 3B = \begin{pmatrix} 6 & 3 \\ 9 & 0 \end{pmatrix}, \quad -A - 3B = -\begin{pmatrix} 3 & 1 \\ 3 & -1 \end{pmatrix} - \begin{pmatrix} 6 & 3 \\ 9 & 0 \end{pmatrix} = \begin{pmatrix} -3-6 & -1-3 \\ -3-9 & 1-0 \end{pmatrix} = \begin{pmatrix} -9 & -4 \\ -12 & 1 \end{pmatrix} $$ 5. Finalmente: $$ x = \frac{3A - B}{2} = \begin{pmatrix} \frac{7}{2} & 1 \\ 3 & -\frac{3}{2} \end{pmatrix}, \quad y = \frac{-A - 3B}{2} = \begin{pmatrix} -\frac{9}{2} & -2 \\ -6 & \frac{1}{2} \end{pmatrix} $$ --- 4. **Problema:** Determinar a matriz inversa de $$ M = \begin{pmatrix} 3 & -1 \\ 4 & -5 \end{pmatrix} $$ **Passos:** 1. Calcular o determinante: $$ \det(M) = 3 \times (-5) - (-1) \times 4 = -15 + 4 = -11 $$ 2. A matriz inversa é dada por: $$ M^{-1} = \frac{1}{\det(M)} \begin{pmatrix} -5 & 1 \\ -4 & 3 \end{pmatrix} = \frac{1}{-11} \begin{pmatrix} -5 & 1 \\ -4 & 3 \end{pmatrix} = \begin{pmatrix} \frac{5}{11} & -\frac{1}{11} \\ \frac{4}{11} & -\frac{3}{11} \end{pmatrix} $$ --- 5. **Problema:** Resolver a equação matricial determinantal $$ \det \begin{pmatrix} 1 & 3 & 2 \\ 2 & 6 & x \\ -3 & 4 & 6 \end{pmatrix} = 0 $$ **Passos:** 1. Calcular o determinante pela regra de Sarrus ou expansão por cofatores: $$ \det = 1 \times \det \begin{pmatrix} 6 & x \\ 4 & 6 \end{pmatrix} - 3 \times \det \begin{pmatrix} 2 & x \\ -3 & 6 \end{pmatrix} + 2 \times \det \begin{pmatrix} 2 & 6 \\ -3 & 4 \end{pmatrix} $$ 2. Calcular os menores: $$ \det \begin{pmatrix} 6 & x \\ 4 & 6 \end{pmatrix} = 6 \times 6 - 4 \times x = 36 - 4x $$ $$ \det \begin{pmatrix} 2 & x \\ -3 & 6 \end{pmatrix} = 2 \times 6 - (-3) \times x = 12 + 3x $$ $$ \det \begin{pmatrix} 2 & 6 \\ -3 & 4 \end{pmatrix} = 2 \times 4 - (-3) \times 6 = 8 + 18 = 26 $$ 3. Substituir: $$ \det = 1(36 - 4x) - 3(12 + 3x) + 2(26) = 36 - 4x - 36 - 9x + 52 = (36 - 36 + 52) - (4x + 9x) = 52 - 13x $$ 4. Igualar a zero e resolver para $x$: $$ 52 - 13x = 0 \Rightarrow 13x = 52 \Rightarrow x = 4 $$ --- **Resposta final:** 1. $M = \begin{pmatrix} 1 & 5 & 10 \\ 7 & 3 & 19 \end{pmatrix}$ 2a. $A - 2B^T = \begin{pmatrix} 0 & 1 \\ 3 & -1 \\ -2 & -6 \end{pmatrix}$ 2b. $(A + B^T)(A^T - B) = \begin{pmatrix} 4 & 7 & -9 \\ 5 & 5 & -15 \\ 1 & -1 & -5 \end{pmatrix}$ 3. $x = \begin{pmatrix} \frac{7}{2} & 1 \\ 3 & -\frac{3}{2} \end{pmatrix}, y = \begin{pmatrix} -\frac{9}{2} & -2 \\ -6 & \frac{1}{2} \end{pmatrix}$ 4. $M^{-1} = \begin{pmatrix} \frac{5}{11} & -\frac{1}{11} \\ \frac{4}{11} & -\frac{3}{11} \end{pmatrix}$ 5. $x = 4$