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Question Number 101645 by mathocean1 last updated on 03/Jul/20

E i s a v e c t o r i a l p l a n e i n B = (\vec{i}, \vec{j})

b a s e . f i s a n e n d o m o r p h i s m o f E .

f (\vec{i}) = 4 \vec{i} - \vec{j} a n d f (\vec{j}) = 2 \vec{i} + \vec{j} .

\vec{u} = x \vec{i} + y \vec{j} \in E a n d x, y \in R .

1) D e t e r m i n a t e f^{- 1} (u) .

Answered by mathmax by abdo last updated on 03/Jul/20

M (f, B) = (\begin{matrix} 4 2 \\ - 1 1 \end{matrix}) we have u (\begin{matrix} x \\ y \end{matrix})

f^{- 1} (u) \to M^{- 1} . u let find M^{- 1} M is root of (4 - u) (1 - u) + 2 = 0 \Rightarrow

(u - 4) (u - 1) + 2 = 0 \Rightarrow u^{2} - 5 u + 6 = 0 \Rightarrow M^{2} - 5 M + 6 I = 0 \Rightarrow

M^{2} - 5 M = - 6 I \Rightarrow - \frac{1}{6} M . (M - 5 I) = I \Rightarrow M^{- 1} = - \frac{1}{6} (M - 5 I)

= (\begin{matrix} - \frac{2}{3} - \frac{1}{3} \\ \frac{1}{6} - \frac{1}{6} \end{matrix}) + (\begin{matrix} \frac{5}{6} 0 \\ 0 \frac{5}{6} \end{matrix}) = (\begin{matrix} - \frac{1}{6} - \frac{1}{3} \\ \frac{1}{6} \frac{2}{3} \end{matrix})

let f^{- 1} (u) = (\begin{matrix} x^{^{'}} \\ y^{^{'}} \end{matrix}) \Rightarrow (\begin{matrix} x^{^{'}} \\ y^{^{'}} \end{matrix}) = M^{- 1} . u = (\begin{matrix} - \frac{1}{6} - \frac{1}{3} \\ \frac{1}{6} \frac{2}{3} \end{matrix}) . (\begin{matrix} x \\ y \end{matrix})

= (\begin{matrix} - \frac{x}{6} - \frac{y}{3} \\ \frac{x}{6} + \frac{2 y}{3} \end{matrix})