## COMMENTS AND SOLUTIONS FOR EXAM 1

```1. (b), (c), and (e) are true;  the others are false.
Note that projections always have rank 1,  and that
the composition of two linear transformations
is always linear (and its matrix is the product
of the matrices).

2.  The initials NTSA on your paper mean that you did
not use the standard algorithm -- there was no
deduction for that.

The standard algorithm goes this way:

Multiply the first row by 1/2 to get:

1 2  3  4
1 2  4  6
3 6 10 14

Subtract the first row from the second to get:

1 2  3  4
0 0  1  2
3 6 10 14

Subtract 3 times the first row from the third to get:

1 2 3 4
0 0 1 2
0 0 1 2

Subract three times the second row from the first to get:

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

Subtract the second row from the third to get:

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

(b) There are two leading 1's  (two nonzero rows),  so the
rank is 2.

(c) The leading variables are x and z  while  y  is free,
and we may solve to get  x = -2 -2y,  z = 2.  The solution
set is therefore

-2 - y
{     y     :  y in R }
2

Our R^3  is  3 by 1 column vectors.  There was no penalty for
the use of the notation  (-2-y, y, 2),  however.  Also, note
that in set notation it must be specified how   y  is allowed
to vary, and that the condition on  y  is separated from
the vector by a colon  :  or a vertical line  | .  Do not
use a comma or semi-colon.  Also, do not put an extra
column vector

x
y
z

inside the set notation.  It is completely correct if you replace
"y" by another variable such as "s"  or "t":  in set notation,
it is a "dummy variable".

(d) The system referred to in (c) is the system given by  A (there
is no reference to the reduced row echelon form) ---  your
answer should use the columns of  A,  not of rref(A) (nor
rref(A) with its 0 row deleted).  The system you write down
from  rref(A)  has the same solutions as the original system
referred to, but it is a different system.   (One point
deduction for using  rref(A).)   The scalars  x, y, z  should
be written to the left of the column vectors:  there was no
penalty  for putting them on the right, however.

2       4         6      8
x  1  + y  2   +  z  4  =   6
3       6        10     14

3.
(a) (1 + 2^2 + 2^2)^(1/2) = 9^(1/2) = 3

1/3
(b) Divide the vector by 3  to get  u =   2/3
2/3
(c) The projection of \$v\$ will be (v.u)u.  The dot product is
6(1/3) + 0(2/3) + 3(2/3) = 2+2 = 4,  and so the projection is
4/3
4u =   8/3
8/3

Typically, anyone who made the mistake of thinking the dot product
is a vector lost all credit on parts (c), (d), (e).  There was a
substantial penalty for using the original vector

1
2
2

1/3
2/3
2/3

In the formula,  (v.u)u   some people used the correct value
of  u  in one spot and the wrong value in another.  This also
incurred a substantial loss of credit that tended to carry through
(c), (d), (e).

(d)  Here,  (v.u)  is  x/3 + 2y/3 + 2z/3 and multiplying by u
gives

x/9 + 2y/9 + 2z/9
2x/9 + 4y/9 + 4z/9
2x/9 + 4y/9 + 4z/9

(e) Thus, the matrix is

1/9 2/9 2/9
2/9 4/9 4/9
2/9 4/9 4/9

The answer should not involve variables.  Little or no credit was
given for matrices coming from thoroughly incorrect answers to part
(d).  Some people thought that the intention here was to generalize
to the case of a line in the direction of an arbitrary unit
vector  u =

a
b
c

where a^2 + b^2 + c^2 = 1.

In that generality, the correct matrix is

a^2  ab  ac
ab b^2  bc
ac  bc c^2

for which full credit was given. (If one does not assume that
the length is one, one should divide all entries by the square
of the length, i.e. by a^2 + b^2 + c^2.)  If you are not sure

4.
(a) (1) 30 degree rotation counterclockwise
(2) dilation by a factor of 3
(3) shear parallel to the e_1 - axis
(b) The image of the unit square is a parallelogram with edges

2
1

and

1
3

and T(e_1/2 + e_2/2) = T(e_1)/2 + T(e_2)/2  will correspond
to the center of the parallelogram, which is the midpoint of
the line joining  T(e_1)  to  T(e_2).  Quite a few people did
not answer the question fully:  they did not show the
parallelogram which is the image of the unit square.

5.
(a)
We row reduce the matrix

1 0 2  1 0 0
0 2 0  0 1 0
1 0 k  0 0 1

Subtract the first row from the last.  Divide the second
row by 2. This gives:

1 0  2    1  0 0
0 1  0    0 1/2 0
0 0 k-2  -1  0 1

It is now clear that the matrix will be invertible (rank 3)
if and only if  k is NOT 2.  Assuming this,  divide the third
row by  k-2:

1 0 2      1      0     0
0 1 0      0     1/2    0
0 0 1  -1/(k-2)   0  1/(k-2)

Finally,  subtract twice the third row from the first:

1 0 0  1 + 2/(k-2)   0  -2/(k-2)
0 1 0       0       1/2     0
0 1 0    -1/(k-2)    0   1/(k-2)

Thus, the inverse is

1 + 2/(k-2)   0  -2/(k-2)
0       1/2     0
-1/(k-2)     0   1/(k-2)

The entry on the upper left is equal to  k/(k-2).   Either