**MATH1025
AF**

**FW00F**

**Answers
to Tutorial 6**

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 (k**u**).**v** = k(**u**.**v**)

k**u**
= (10, 0, -15, 5); (k**u**).**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)^{2}t^{2}
+ 44^{2}t^{2 }+ (-6)^{2}t^{2} + 11^{2}t^{2}
= 1 or t^{2} = 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.

14. 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 R^{2}. By inspection
or by solving the linear equations, we can see that any vector **w**
in the range must have the property that w_{1}
= 2w_{2}. 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(c**u**)
= cT(**u**). Now **u** + **v **= (x+a, y+b) and c**u**
= (cx, cy).

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

(b)
T(**u** + **v**) = ((x+a)^{2}, y+b) = (x^{2} +
2xa + a^{2}, y + b) which is **not** equal to

T(**u**)
+ T(**v**) = (x^{2}, y) + (a^{2}, 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(c**u**) = (-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(c**u**) = (cx, 0) = c(x, 0) = cT(**u**). Thus T
is a linear transformation.

13.
We use the basis vectors **e**_{1} and **e**_{2}
which are the unit vextors along the x- and y-axes.

See answers in text.

21. We have to show that Thm 4.3.2 holds as in question 8.

(a)
If T maps any vector into the zero vector then T(**u** + **v**)
= (0, 0) = T(**u**) + T(**v**) and T(c**u**) =

(0,
0) = cT(**u**). Thus this transformation is linear.

(b)
If T maps any vector into the vector (1, 1) then T(**u** + **v**)
= (1, 1) but T(**u**) + T(**v**) = (1, 1) +

(1,
1) = (2, 2). Thus this transformation is **not **linear.