Explanation Given: The mapping G ( u , v ) = ( 2 u + v , 5 u + 3 v ) Calculation: Since G ( u , v ) = ( 2 u + v , 5 u + 3 v ) ........ ( 1 ) Therefore, G ( G − 1 ( x , y ) ) = G ( 3 x − y , − 5 x + 2 y ) Here u = 3 x − y , v = − 5 x + 2 y .substitute u and v in relation (1) G ( 3 x − y , − 5 x + 2 y ) = ( 2 ( 3 x − y ) + ( − 5 x + 2 y ) , 5 ( 3 x − y ) + 3 ( − 5 x + 2 y ) ) G ( 3 x − y , − 5 x + 2 y ) = ( 6 x − 2 y + ( − 5 x + 2 y ) , 15 x − 5 y + ( − 15 x + 6 y ) ) G ( 3 x − y , − 5 x + 2 y ) = ( 6 x − 5 x , − 5 y + 6 y ) G ( 3 x − y , − 5 x + 2 y ) = ( x , y ) Hence G ( G − 1 ( x , y ) ) = ( x , y ) Now it only remains to show that G − 1 ( G ( u , v ) ) = ( u , v ) since G − 1 ( x , y ) = ( 3 x − y , − 5 x + 2 y ) ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.6, Problem 9E

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The inverse of $G$ is $G^{- 1} (x, y) = (3 x - y, - 5 x + 2 y)$ .

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Chapter 16.6, Problem 8E

Chapter 16.6, Problem 10E

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Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 5PQ Ch. 16.1 - Prob. 6PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E

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The inverse of G is G − 1 ( x , y ) = ( 3 x − y , − 5 x + 2 y ) .

Solution Summary: The author explains that the inverse of G is G 1 ( x, y ) = ( 3.

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The inverse of $G$ is $G^{- 1} (x, y) = (3 x - y, - 5 x + 2 y)$ .

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Chapter 16 Solutions

The inverse of G is G − 1 ( x , y ) = ( 3 x − y , − 5 x + 2 y ) .

Solution Summary: The author explains that the inverse of G is G 1 ( x, y ) = ( 3.

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To describe: The inverse of Gis G−1(x,y)=(3x−y,−5x+2y).

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Chapter 16 Solutions

To describe:

The inverse of $G$ is $G^{- 1} (x, y) = (3 x - y, - 5 x + 2 y)$ .