For typographical reasons, brackets around matrices are omitted here.

Section 1.3

4.

To find the rank of the matrix A, put it in RREF:

1 4 7

2 5 8

3 6 9

1 4 7

0 -3 -6

0 -6 -12

1 4 7

0 1 2

0 -6 -12

1 0 -1

0 1 2

0 0 0

The rank is 2.

30.

There must be three leading 1's in different columns
moving to the right and these must occur in the first
three rows. So the first three rows of the 5 by 3 matrix
look just like a 3 by 3 identity matrix. The last two
rows must be 0. So the only possible row reduced echelon
form for a 5 by 3 matrix of rank 3 is

1 0 0

0 1 0

0 0 1

0 0 0

0 0 0

36.

1 4 7

2 5 8

3 6 9

(the same as in problem 4. above)

is the only matrix taking the specified values on e_1, e_2, e_3.

Section 2.1

6.

The value of T on

x_1

x_2

is

x_1 + 4x_2

2x_1 + 5x_2

3x_1 + 6x_2

and this is the same as A times the vector

x_1

x_2

where A =

1 4

2 5

3 6

This IS a linear transformation.

10.

To invert the matrix

1 2

4 9

we want to solve the system

x_1 + 2x_2 = y_1

4x_1 + 9x_2 = y_2

for x_1, x_2 in terms of y_1, y_2, if possible.

Subtract 4 times the first row from the second:

x_1 + 2x_2 = y_1

x_2 = -4y_1 + y_2

Subtract twice the second row from the first:

x_1 = 9y_1 - 2y_2

x_2 = -4y_1 + y_2

This shows that the original matrix is invertible with inverse:

9 -1

-4 1

24.

The value of the transoformation on e_1 is e_2 and
the value on e_2 is -e_1. This means
that the transformation is a rotation, 90 degrees
counterclockwise. The face shown in the book is rotated
90 degrees counterclockwise: the eyes wind up in the
second and third quadrants while the mouth stretches over
the first and fourth quadrants.

Section 2.2

2.

Just substitute 60 degrees for alpha in the formula given
in Fact 2.2. This produces

cos(60) -sin(60)

sin(60) cos(60)

Here, cos(60) = 1/2, and sin(60) is the square root of 3 divided by 2.