44_hwc1SolnsODDA

# 44_hwc1SolnsODDA - ± u v ² = ± 1 2 3 ²± x y ² , ± x...

This preview shows page 1. Sign up to view the full content.

6.1 Let A be the matrix A = h 1 2 0 3 i . Find the equation describing the image of S under the linear transformation corresponding to A , and graph this image of S , where: (a) S is the line y –axis. (b) S is the line x + y = 3. (c) S the unit circle. SOLUTION To ﬁnd an equation describing the image of a set S under an invertible transformation, we can proceed just as we did in Example 34 – at least whenever S itself is described by an equation. This is the case in this exercise. (Recall that the equation describing the y –axis is x = 0.) The procedure from Example 34 is to invert the transformation, so that we ﬁnd formulas for x and y in terms of u and v , and then to substitute these formulas into the equation describing S . Using the formula for inverting 2 × 2 matrices; that is, formula (3.19) in Theorem 7, we ﬁnd ± 1 2 0 3 ² - 1 = 1 3 ± 3 - 2 0 1 ² . Therefore, when
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ± u v ² = ± 1 2 3 ²± x y ² , ± x y ² = 1 3 ± 3-2 1 ²± u v ² = 1 3 ± 3 u-2 v v ² . That is, x = (3 u-2 v ) / 3 and y = v/ 3 . ( * ) Now it is easy: We just substitute these expressions into the three equations for the three given sets S . (a) The equation for the y-axis is x = 0. Using ( * ) to express this in terms of u and v , we get (3 u = 2 v ) = 0, or, what is the same, v = 3 2 u . (This the the equation of the line through the origin in the u, v plane with slope 3 / 2.) (b) As in part (a) , we use ( * ) to express the equation for S , here x + y = 3, in terms of u and v . The result is: (3 u-2 v ) / 3 + v/ 3 = 3 , which simpliﬁes to v = 3 u-9 , 21 /september/ 2005; 22:09 45...
View Full Document

## This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

Ask a homework question - tutors are online