MATH1025 AF

FW00F

Section 4.1

Questions 1, 5 - see answers in text

13. part (b) of Thm 4.1.2 - demonstrate (u + v).w = u.w + v.w

u.w = -5; v.w = 5; u.w + v.w = 0

u + v = (6, 0, 0, 6); (u + v).w = 0

part (c) of Thm 4.1.2 - demonstrate (ku).v = k(u.v)

ku = (10, 0, -15, 5); (ku).v = 20

u.v = 4; k(u.v) = 20

16. Let the unknown vector be s = (a, b, c, d). Then if this vector is orthogonal to the other 3 we have s.u = 2a + b - 4c = 0

s.v = -a - b + 2c + 2d = 0

s.w = 3a + 2b + 5c + 4d = 0

Solving this system of linear equations by gaussian elimination we get (interchanging R1 and R2)

Let d = 11t (to get rid of fractions). Then c = -6t, b = 44t, a = -34t. But if s has norm 1 then

(-34)2t2 + 442t2 + (-6)2t2 + 112t2 = 1 or t2 = 1/3249 and t = 1/57 or -1/57 giving

s = 1/57(-34, 44, -6, 11) or -1/57(-34, 44, -6, 11) as the two vectors

Section 4.2

Questions 1, 2, 6 - see answers in text

9. Use standard matrices given in Table 3 of this section - see answers in text.

11. Use standard matrices given in Table 5 of this section - see answers in text.

12. (a) Standard matrix is - see answer in text.

(b) Standard matrix is - see answer in text.

(c) Standard matrix is - see answer in text.

(d) Standard matrix is - see answer in text.

15. (a) Standard matrix is - see answer in text.

(b) Standard matrix is - see answer in text.

(c) Standard matrix is - see answer in text.

16. (a) Using the standard matrices from question 12 (d) and Table 2 the composition is

which is a reflection about the x-axis.

(b) Using the standard matrices from Table 4 and Table 8 the compositions is

(c) Using the standard matrices from Table 2 and Table 8 the compositions is

20. (a) Using the standard matrices from Table 4 the compositions are

so the composition is commutative.

(b) Using the standard matrix from Table 6 the compositions are

so the composition is commutative.

(c) Using the standard matrices from Table 4 and Table 6 the compositions are

so the composition is not commutative.

Section 4.3

Question 2 - see answers in text - the transformation is one-to-one if the determiniant of the standard matrix is non-zero.

3. The standard martrix for the transformation iswhose determinant is zero. Thus the transformation is not one-to-one and by Thm 4.3.1 the range cannot be all of R2. By inspection or by solving the linear equations, we can see that any vector w in the range must have the property that w1 = 2w2. Hence any vector without this property, say (1, 1) will not be in the range of the linear transformation.

5. (a) The standard matrix is which has a determinant of 3 so the transformation is one-to-one. See answers in text for inverse.

(b) The standard matrix is which has a determinant of 0 so the transformation is not one-to-one and no inverse exists.

(c) The standard matrix is which has a determinant of -1 so the transformation is one-to-one. See answers in text for inverse.

(d) The standard matrix is which has a determinant of 0 so the transformation is not one-to-one and no inverse exists.

8. Let u = (x, y) and v = (a, b). Then we must show that T(u + v) = T(u) + T(v) and

T(cu) = cT(u). Now u + v = (x+a, y+b) and cu = (cx, cy).

(a) T(u + v) = (2(x+a), y+b) = (2x,y) + (2a, b) = T(u) + T(v); T(cu) = (2cx, cy) = c(2x, y) = cT(u). Thus T is a linear transformation.

(b) T(u + v) = ((x+a)2, y+b) = (x2 + 2xa + a2, y + b) which is not equal to

T(u) + T(v) = (x2, y) + (a2, b) so T is not a linear transformation.

(c) T(u + v) = (-(y+b), x+a) = (-y, x) + (-b, a) = T(u) + T(v); T(cu) = (-cy, cx) = c(-y, x) = cT(u). Thus T is a linear transformation.

(d) T(u + v) = (x+a, 0) = (x, 0) + (a, 0) = T(u) + T(v); T(cu) = (cx, 0) = c(x, 0) = cT(u). Thus T is a linear transformation.

13. We use the basis vectors e1 and e2 which are the unit vextors along the x- and y-axes.