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Recall V = 0 (V3.13) and consider this as a vector space over F = Q(v2) Show that 1, V3 is a basis.

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- solution :- V = 0(2,13) = {a +bla a+bJa+CJ3 +0 J6; abstras F = 0(52)={xty Je, ny far B = { 1,337 Basis for veckor space v dB is finearly Independent. det HY EF = 0 (52) such that xil + y - 33 = 0 x + a re , di, Xfo How det x y = y + Y J2 y , co treSo Set B is finearly Independent, So set B is Basis foo vedo space v fiepe F Answer over

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