Exercise 3.F.13

Answers

Proof. _(a)_

\begin{aligned} T^{'} (φ_{1}) (x, y, z) & = φ_{1} \circ T (x, y, z) \\ = φ_{1} (4 x + 5 y + 6 z, 7 x + 8 y + 9 z) \\ = 4 x + 5 y + 6 z \end{aligned}

\begin{aligned} T^{'} (φ_{2}) (x, y, z) & = φ_{2} \circ T (x, y, z) \\ = φ_{2} (4 x + 5 y + 6 z, 7 x + 8 y + 9 z) \\ = 7 x + 8 y + 9 z \end{aligned}

_(b)_ Since $ψ_{1} (x, y, z) = x$ , $ψ_{2} (x, y, z) = y$ and $ψ_{3} (x, y, z) = z$ , substituting these in the results from item _(a)_, we get

T^{'} (φ_{1}) (x, y, z) = 4 ψ_{1} (x, yz) + 5 ψ_{2} (x, y, z) + 6 ψ_{3} (x, y, z)

Thus $T^{'} (φ_{1}) = 4 ψ_{1} + 5 ψ_{2} + 6 ψ_{3}$ . Similarly, we get $T^{'} (φ_{2}) = 7 ψ_{1} + 8 ψ_{2} + 9 ψ_{3}$ . □

celiopassos

2017-10-06 00:00