Exercício Geometria Analítica Resolvida-2022 1

C = \left\{ \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right\} (i) C \subseteq I \alpha \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} + \beta \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} + \gamma \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha & -\beta \\ \alpha + \beta & \gamma \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \left\{ \begin{array}{l} \alpha = 0 \\ -\beta = 0 \\ \alpha + \beta = 0 \\ \gamma = 0 \end{array} \right. \alpha = \beta = \gamma = 0 Portanto, C \subseteq I (ii) C gera U a\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} + b\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} + c\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x & y \\ z & t \end{bmatrix} \begin{bmatrix} a & -b \\ a + b & c \end{bmatrix} = \begin{bmatrix} x & y \\ z & t \end{bmatrix} \left\{ \begin{array}{l} x = a \\ y = -b \\ z = a + b \\ t = c \end{array} \right. \begin{bmatrix} a = x \\ b = -y \\ c = t \end{bmatrix} Portanto, C gera U, logo, é uma base 5b B = \left\{ \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right\} C = \left\{ \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right\} \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = a \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} + b \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} va = 0, b = 1, c = 0 \text{ (imediato)} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = e \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} + f \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} + g \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} e + f & e \\ f & g \end{bmatrix} q = 0, e = -1, f = 1 \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = h \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} + i \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} + j \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} h = 0, i = 0, j = 1 ( 2 ) A = [ 1 1 -1 0 ] [ 0 1 0 0 ] [ -1 0 2 0 ] [ 1 1 1 -1 ] det A = 1 * (-1) .. [ -1 -1 0 | 1 0 ] [ -1 2 0 | -1 2 0 ] [ 1 -1 1 | 1 -1 ] 1 * (-1)^4 [ -2 + 1 ] = -1 Como det A ≠ 0, a matrix é inversível Encontramos a inversa utilizando a eliminação de Gauss [ 1 1 -1 0 | 1 0 0 0 ] L1 [ 0 1 0 0 | 0 1 0 0 ] L2 [ -1 0 2 0 | 0 0 1 0 ] L3 [ 1 1 1 -1 | 0 0 0 1 ] L4 L3 -> L1 + L3 [ 1 1 -1 0 | 1 0 0 0 ] [ 0 1 0 0 | 0 1 0 0 ] [ 0 1 0 | 1 1 0 1 ] [ 1 1 1 -1 | 0 0 0 1 ] L4 -> -L1 + L4 [ 1 1 -1 0 | 1 0 0 0 ] [ 0 1 0 0 | 0 1 0 0 ] [ 0 1 0 | 1 -1 0 1 ] [ 0 0 2 -1 | -1 0 0 1 ] L3 -> -L2 + L3 [ 1 1 -1 0 | 1 0 0 0 ] [ 0 1 0 0 | 0 1 0 0 ] [ 0 0 1 0 | 1 -1 1 0 ] [ 0 0 2 -1 | -1 0 0 1 ] L4 -> -2L3 + L4 [ 1 1 -1 0 | 1 0 0 0 ] [ 0 1 0 0 | 0 1 0 0 ] [ 0 0 1 0 | 1 -1 1 0 ] [ 0 0 0 -1 | -3 2 -2 1 ] L4 -> -L4 [ 1 1 -1 0 | 1 0 0 0 ] [ 0 1 0 0 | 0 1 0 0 ] [ 0 0 1 0 | 1 -1 1 0 ] [ 0 0 0 1 | 3 -2 2 -1 ] L1 -> L3 - L1 [ 1 1 0 0 | 2 -1 1 0 ] [ 0 1 0 0 | 0 1 0 0 ] [ 0 0 1 0 | 1 -1 1 0 ] [ 0 0 0 1 | 3 -2 2 -1 ] L1 -> -L2 + L1 [ 1 0 0 0 | 2 -2 1 0 ] [ 0 1 0 0 | 0 1 0 0 ] [ 0 0 1 0 | 1 -1 1 0 ] [ 0 0 0 1 | 3 -2 2 -1 ] A^-1 = [ 2 -2 1 0 ] [ 0 1 0 0 ] [ 1 -1 1 0 ] [ 3 -2 2 -1 ] 3) \vec{a} = (3, 1, m) \vec{b} = (m, 12, 5, 2) \vec{c} = (2m, 8, m) \vec{a} + \vec{b} = (m+4, -4, m+2) \vec{c} - \vec{a} = (2m-2, 7, 0) (\vec{a} + \vec{b}).(\vec{c} - \vec{a}) <0 (nao ortogonais) (m+4)(2m-2) -28 =0 2m^2 - 2m + 8m - 8 - 28 = 0 2m^2 + 6m - 36 = 0 m^2 + 3m - 18 = 0 S = -3 P = -18 m = -6 m = 3 m = -6 ou m = 3 4) \begin{cases} y = 2x - 3 \\ y = -x + 4 \end{cases} \alpha x + \beta y - z - 2 = 0 Substituindo na equacao do plano, temos \alpha x + \beta (2x - 3) + x - 4 - 2 = 0 \alpha x + 2\beta x - 3\beta + x - 6 = 0 (\alpha + 2\beta + 1)x - 3\beta - 6 = 0 \begin{cases} \alpha + 2\beta + 1 = 0 \\ -3\beta - 6 = 0 \end{cases} -3\beta = 6 \beta = -2 \alpha - 4 + 1 = 0 \alpha - 3 = 0 \alpha = 3 a) 5) x = (y + z)f \begin{bmatrix} y+z \\ y \\ z \end{bmatrix} = y \begin{bmatrix} 1 \ \ 0 \end{bmatrix} + z \begin{bmatrix} 1 \ \ 0 \end{bmatrix} Portanto, os elementos de B geram as matrizes do conjunto V. Resta mostrar que o conjunto e LI. \alpha \begin{bmatrix} 1 \ \ 1 \end{bmatrix} + \beta \begin{bmatrix} 1 \ 0 \end{bmatrix} + \gamma \begin{bmatrix} 0 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \begin{cases} \alpha + \beta = 0 \\ \alpha = 0 \\ \beta = 0 \\ \gamma = 0 \end{cases} Como os escalares nao todos nulos, o conjunto e LI. Depois Portanto, B e uma base A matriz de mudanca de base de B para C e A = [0 -1 0 1 1 0 0 0 1] Verificamos se a matriz e diagonalizavel det(A - λI) = 0 = [-λ -1 0 1 λ 0 0 0 1] -λ(1 - λ)^2 + (1 - λ) = 0 (1 - λ) [ -λ(1 - λ) + 1 ] = 0 (1 - λ) [ λ^2 - λ + 1 ] = 0 λ = 1 λ^2 - λ + 1 = 0 Δ = 1 - 4.1.1 Δ = -3 não ha raizes reais Como nem todos os autovalores são reais, segue que a matriz não é diagonalizavel sobre K = ℝ