## Test 1: Solutions, comments, marking scheme (in part)

I am very late getting this file to you; sorry for this.
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1. (a)  { (2+t-s, s, 1+3t, t) | s,t are arbitrary }.

":" and "such that" are also ok in place of "|" above.

OK to write the ordered 4-tuple as a column matrix instead.

OK to write "are in |R" instead of "are arbitrary".

Mark deductions:

x,z for s,t:  0.3
omitting the "|": 0.7
omitting the "|": 1.2
missing commas:   0.2 each
square brackets on 4-tuple:  0.4
errors in entries of 4-tuple: 1 each
missing "set" brackets:  1
round brackets in place of set brackets:  0.7

(b)

This is just a matter of picking two different
pairs of values for the parameters in part (a)
and writing down the two corresponding 4-tuples.

I guess it is possible to get 4 marks out of 4
in part (b) even if (a) is left blank....  I
only asked for two specific solutions; I did not
say people had to use the parametric description
to find their solutions.

2.  First step, OBVIOUSLY, is to sweep under
the leading 1 in row 2, to get a last row that is:

0     0     c^2-16      c+4

Doing this was "worth" 1.5 marks.  People lost 0.3
for not labelling the elem. row op. according to one
of the two approved notations from class.

Next step in GJ algorithm is clearly to get a leading
1 in the third row.  Where this 1 sits depends on whether
c^2 - 16  is  0  or not.

If  c  is not   4 or -4,  then the system has a unique
solution.

(Reasons:  The coeff. matrix of the system is invertible.

Or:  There is no leading 1 in column 4 of the red. r-e.
form of the augmented matrix of the system (so it is
consistent), and no non-leading variable, hence no
parameters in a description of the solution set.)

0.5 marks for noting that c not equal to 4 or -4  is an
important case to consider.

1 mark for saying the stuff about unique solution.

2 marks for giving one of the two reasons above.

If  c=4  (0.5 marks for realizing this is an important case),

the system has no solutions. (1 mark for saying this)

Reasons:  The GJ form of M has a leading 1 in the last column.
(2 marks for this or something tantamount to this.)

If  c is -4,  the system has infinitely many solutions.
(0.5 for identifying this as an important case; 1 for
stating the above)

Reasons:  GJ(M) then has no leading 1 in the last column,
so the system is consistent, but also there is a non-leading
variable (i.e., a parameter) in the description of the solution
set.  (2 marks)

3. 0 marks if anything but c is circled.
c must be circled.  Reason:  Something like this:

Every lin. system has either no solutions, or exactly
one solution, or infinitely many solutions.

Or, the above, with a comma at the end, then "so if
the system has at least two solutions, then it must
have infinitely many."

Deductions:

0.2 for thinking there is an English word "infinetly".
If you want the correct spelling, turn to the previous
page of the test and copy it down....

0.8 for saying "infinite" instead of "infinitely many".
No n-tuple is infinite.

0.2 for calling a system a matrix.  This is disgusting.

0.6 for claiming at the start that a 4X7 system must have
3 parameters, at the start.  If the system is inconsistent,
the solution set is empty.  So it is hard for there to be
parameters in its description....

1.5 for the same remark but "2 parameters"....

1.1 for writing down all of the above but not circling c!
(Instead, writing "none of the above".)

0.3 for each sentence in excess of the one asked for.

0.8 for adding a fifth choice "e  infinitely many".

Deductions were also made for lousy English.

This is a bonehead, dirt simple problem.  5 free marks.
I even promised in class to ask such a problem.

4.  Let  X  and  Z  be rows 1, 2 of the matrix on
the far left.  Then  XA = ZA,  because these are the first
two rows of the matrix on the right, and those are equal.
(2.2 marks for this much)

So if we let  Y = X - Z,  then  YA = (X-Z)A = XA - ZA = 0.
(2 marks for this)

But  Y is not 0.  So  A  is not invertible.  (1.8 marks)

Comments:  We came across this characterization of inverti-
bility in class at least three times.  The Suggestion
in the problem makes this problem easy.  I think at most
one person out of 19 got it.

I gave 1/6 for some sense involving determinants.

Another way:

[x y z]A = [1 3 3]  is a system having two solutions.

So  + 1 +         T      T  + x +
| 3 |  =  (YA)   =  A   | y |   has at least two
+ 3 +                   + z +
T
solutions too.  So by a result from class,  A  is not invertible.
So  A  is also not invertible  (we saw an exercise in class acc.
to which if A is invertible, then so is its transpose, and the
inverse of the transp. of A is the transp. of the inverse of A).

5.

A problem JUST like this was done in class.  This was free
marks.

6. (a)  First blank:  HD = DH = I  . (1.8 marks)

Second blank:  there is such an H         . (1.2 marks)

(b)  Call the displayed equation (*).  There are various
correct ways of writing up a proof.  Many of them involve the
fact that we used several times in class that the inverse of
a product is the product of the inverses, taken in the reverse
order.  One can also do various parts of the proof just by
using the DEFINITION of left/right inverse of a square
matrix and computing various products directly.  Here I was
not so much interested in checking sizes at the beginning as
in having the key ideas for the proof.

Pf:

(*) holds

iff

A^{-1} (A^{-1} + B)^{-1}  =  (A + B^{-1})^{-1} B^{-1}
----------            ----------
Call this C.          Call this D.

iff

(CA) inverse  =  (BD) inverse

iff  (just crank out CA and BD!  :-) )

(I + BA) inverse  =  (BA + I) inverse.

But  +  is commutative.  So the last line holds.  So (*) holds. //

The grades out of 50 were all written on the back of the
test papers.

The circled (i.e. the higher) number is the grade I recorded
out of 50, in each case.

```

The Chief