1) Ache a transformação linear T R³? R²
tal que t (1,0,0) = (2,0),T(0,1,0) = (1,1) e T(0,0,1) = (0,-1)
R : [ 1,0,0) , (0,1,0) ,(0,0,1) ] base R³
seja (x,y,z) ? R³
T(1,0,0) = (2,0)
T(0,1,0) = (1,1)
T(0,0,1) = (0,-1)
(x,y,z) = a(1,0,0) + b (0,1,0) + c (0,0,1)
(x,y,z) = (a,0,0) +(0,b,0) + (0,0,c) = (x,y,z) (a,b,c)
x=a
y=b
z=c
(x,y,z) = x(1,0,0) + y(0,1,0) +z(0,0,1)
T(x,y,z) = xT(1,0,0)+yT(0,1,0)+zT(0,0,1)
T(x,y,z) = x(2,0)+y(1,1) +z(0,-1)
T(x,y,z) = (2x,0)+(y,y) +(0,-z)
T(x,y,z) = (2x+y,y-z)
essa questao esta certa
![{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[5]}](/latexrender/pictures/19807748a214d3361336324f3e43ea9a.png "{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[5]}")
![{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[2]{5}}](/latexrender/pictures/3d7908e5b4e397bf635b6546063d9130.png "{(0,05)}^{-\frac{1}{2}}=\frac{10}{\sqrt[2]{5}}")
, ou seja, 1 dividido por 20 é igual a 0.05 . Sendo assim, a função final é igual a vinte elevado à meio. ![{0,05}^{-\frac{1}{2}} = {\frac{1}{20}}^{-\frac{1}{2}} = {\frac{20}{1}}^{\frac{1}{2}} = \sqrt[2]{20}](/latexrender/pictures/c0100c6f4d8bdbb7d54165e6be7aff04.png "{0,05}^{-\frac{1}{2}} = {\frac{1}{20}}^{-\frac{1}{2}} = {\frac{20}{1}}^{\frac{1}{2}} = \sqrt[2]{20}")
da seguinte forma:
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da seguinte forma:
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