A/

1)	2)
3) inconsistent, no solutions	4)

.

5) There are nontrivial solutions when k = 4 or k = 2.

.

6) The system matrix reduces to , so:

a) if a = – 4, a² – 16 = 0 but a – 4 = –8 so threre will be no solutions.

b) if a is not = – 4 there is a unique solution.

c) If a = 4 there are infinite solutions.

.

B/

1)

2)

3)

4)

.

5)

.

6) Proof:

Let A be an m × n matrix and B be an r × q matrix.
Since AB exists, n must be equal to r, so AB is an m × q matrix.
Now, BA also exists, therefore q must be equal to m, so BA is an r × n matrix.
But, m = q, therefore AB is a square matrix.
And r = n, therefore BA is a square matrix.

.

C/

1)

2)

3)

4)

.

5) The system matrix .

Since row 3 (left of the bar) is 3 times row 2, to be consistent, c – 8a must = 3(b – 2a)
which gives c = 2a + 3b for the system to be consistent.

.

D/

1)

a) det = – 36. (We switched 2 rows ( –1), then multiplied by –2 and 3.)

b) det = – 6. (We switched 2 rows ( –1), then multiplied by – 1.)

c) det = 18. (We multiplied a row by – 3.)

.

2) .

.

E/

1) a)

b)

c) cos A = – ½

d)

.

2) The parametric equations give us the point P (0, 1, –3) and the vector .

The normal n of the plane = the cross product of vector l and the vector joining the 2 points.

So, The equation of the plane is 3x – y – z – 2 = 0.

.

3) The line vector and l $ n = 0, so the line // plane.

.

4) The line of intersection of the planes is: [ x, y, z ] = [ –5 – 2t, –7 – 6t, t ].

So the normal of the plane we seek is

This makes the plane's equation x – y – 4z – 2 = 0.

.

F/

1) a) The system matrix .

The solution therefore is x = 16t, y = 19t and z = t.

So a basis for the nullspace of A is the vector .

b) A basis for the row space of A are the vectors .

c) A basis for the column space of A are the vectors .

d) nullity = dimension of basis for nullspace. nullity = 1

e) rank A = 2

f) no, since the 3rd component must = 0 to be in the column space of A.

g) The nullspace of A is a line through the origin.

.

.

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