Answer
We show that ${G^{ - 1}}\left( {x,y} \right) = \left( {3x - y, - 5x + 2y} \right)$ is the inverse of $G\left( {u,v} \right) = \left( {2u + v,5u + 3v} \right)$.
Work Step by Step
We are given ${G^{ - 1}}\left( {x,y} \right) = \left( {3x - y, - 5x + 2y} \right)$. We show that ${G^{ - 1}}$ is the inverse of $G$.
1. Using $G\left( {u,v} \right) = \left( {2u + v,5u + 3v} \right)$, evaluate $G\left( {{G^{ - 1}}\left( {x,y} \right)} \right)$
$G\left( {{G^{ - 1}}\left( {x,y} \right)} \right) = \left( {2\left( {3x - y} \right) - 5x + 2y,5\left( {3x - y} \right) + 3\left( { - 5x + 2y} \right)} \right)$
$G\left( {{G^{ - 1}}\left( {x,y} \right)} \right) = \left( {x,y} \right)$
2. Evaluate ${G^{ - 1}}\left( {G\left( {u,v} \right)} \right)$
${G^{ - 1}}\left( {G\left( {u,v} \right)} \right) = \left( {3\left( {2u + v} \right) - \left( {5u + 3v} \right), - 5\left( {2u + v} \right) + 2\left( {5u + 3v} \right)} \right)$
${G^{ - 1}}\left( {G\left( {u,v} \right)} \right) = \left( {u,v} \right)$
From the results above, we conclude that ${G^{ - 1}}$ is the inverse of $G$.
