Mathematics 21b Review Problems

Spring 98 Exam 1 Solutions

(1) The equations are

    w +    n +    s = 100
 .6 w + .4 n + .4 s = w
 .3 w + .5 n        = n
 .1 w + .1 n + .6 s = s

and they can be rewritten

    w +    n +    s = 1
-.4 w + .4 n + .4 s = 0
 .3 w - .5 n        = 0
 .1 w + .1 n - .4 s = 0

Solving by Gauss-Jordan gives W = 50, N = 30, and S = 20.

(2)

A is a rotation/dilation matrix, rotating counter-clockwise by 45 degrees and stretching by sqrt(2). So A turns the face into the second quadrant and stretches it a bit.
A^4 is also a rotation/dilation matrix, rotating counter-clockwise by 180 degrees and stretching by 4. So A^4 flips the face over the x-axis and stretches it by 4.
A^4 is a diagonal matrix with -4's along the diagonal.
A^17 is rotation by 765 degrees and dilation by (sqrt(2))^17. So
```
         256   -256
A^17 = 
         256    256
```

(3)

  -1/3   2/3  -2/3        0   0  -1           0   0   -1
B= 2/3  -1/3  -2/3     C= 1   0   0       E= -1   0    0
  -2/3  -2/3  -1/3        0  -1   0           0   1    0


   1/3   1/3  -1/3        0   0  -1   
A= 1/3   1/3  -1/3     D= 0   1   0   
  -1/3  -1/3   1/3        1   0   0

(4) The rref of A is

0  0   2  2
1  2   1  2
2  4  -1  1

So a basis for the image is (0,1,2) and (2,1,-2). A basis for the kernel is (-2,1,0,0) and (-1,0,-1,1).

(5)

```
1  0  0
0  1  0
```
is one possible form.
Not possible, as dim im + dim ker must be 3, but dim V + dim W = 2.
```
 1  0  -1
-1  0   1
```
is one possible form.