1. Consider the following system of linear equations x

1. Consider the following system of linear equations x
1. Consider the following system of linear equations
x − 2y + 3z
=
2
2x − 2y − z
=
1
2x − 4y + 6z
= α
(a) (1 mark) Write the equation in the form of an augmented matrix
(A | b).
Solution:


1 −2 3 | 2
(A | b) =  2 −2 −1 | 1 
2 −4 6 | α
(b) (3 marks) Using Gaussian elimination, find an echelon form of (A | b).
Solution:




1 0 −4 |
−1
1 −2 3 |
2
−3 
−3  →  0 2 −7 |
(A | b) →  0 2 −7 |
0 0 0 | α−4
0 0
0 | α−4
(c) (1 mark) For which values of α does the system of equations have
solution(s)?
Solution: The system is consistent iff α = 4.
(d) (1 mark) Assuming solutions exist, and by considering the rank of
A, how many parameters are in the general solution.
Solution: There is one parameter in the general solution.
(e) (2 marks) Express the general solution in vector form.
Solution: The general solution is



 

x
8
−1
 y  = λ  7  +  −3/2 
z
2
0
for some λ ∈ R.
2. Let A, B ∈ Rn×n . Suppose that A2 = 1 and B T = B. Simplify the
following expressions
−1 −1
(a) (2 marks) AB −1
A
Solution:
−1 −1
AB −1
A = BA−2 = B
T
T
(b) (2 marks) (BA) AB −1
Solution:
T
T
T
(BA) AB −1 = AT B T B −1 AT = I
3. (a) (3 marks) Consider the vectors v1 , v2 , v3 where
 




0
1
1
v1 =  2  v2 =  1  v3 =  −1 
1
−1
2
1
Show that {v1 , v2 , v3 } is a linearly independent set.
Solution: We show that the only solution to λ1 v1 + λ2 v2 + λ3 v3 = 0
is λ1 = λ2 = λ3 = 0. We can write the system of equations in matrix
form, and use Gaussian elimination to solve:





1 1
0 | 0
1 1
0 | 0
1 1
0
 2 1 −1 | 0  →  0 −1 −1 | 0  →  0 −1 −1
1 −1 2 | 0
0 −2 2 | 0
0 0
4
This shows that the only solution is λ1 = λ2 = λ3 = 0.
(b) (1 mark) What is span {v1 , v2 , v3 }
Solution: R3
4. Let
A =
2
1
1
3
3
−1
(a) (2 marks) Find the inverse of A.
Solution:
A
−1
=
1
5
−1
2
(b) (2 marks) Use part a) to solve
2x + y
=
b1
x + 3y
=
b2
Solution:
x
y
=
=
1
5
1
5
3
−1
−1
2
3b1 − b2
2b2 − b1
b1
b2
5. State whether the following subsets W of R3 are subspaces. In each case,
if W is a subspace of R3 , then find a basis for W . If not, then find two
vectors u, v ∈ W such that u + v 6∈ W .
(a) (3 mark) W = (x, y, z) ∈ R3 | xyz = 0
Solution: The subset W ⊂ R3 is not a subspace, since (1, 1, 0),
(0, 1, 1) ∈ W , but their sum (1, 2, 1) is not in W .
(b) (3 mark) W = im (B) = Bx | x ∈ R4 where


1 2 1 2
B =  2 4 3 0 
1 2 1 2
Solution: The subset W is a subspace of R3 . We perform Gaussian
elimination to obtain


1 2 1 2
B →  0 0 1 −4 
0 0 0 0
so a basis for col (A) consists of the first and third columns of B.
2

| 0
| 0 
| 0
For parts a) and b) above, I gave 1 mark for stating whether W is a
subspace. For part a) I gave 2 additional marks for a correct proof that
W is not a subspace. For part b) I gave 1 additional mark for each basis
element of W .
6. Consider the matrix

A
1
=  2
1
2
1
1

4
3 
1
(a) (1 mark) Is A invertible?
Solution: Yes.
(b) (3 marks) If A is invertible, find its inverse. If not, find a linear
relation between the columns of A.
Solution:


1 2
4 | 1 0 0
(A | I3 ) →  0 −3 −5 | −2 1 0 
0 −1 −3 | −1 0 1


1 2
4 | 1 0 0
→  0 −1 −3 | −1 0 1 
0 −3 −5 | −2 1 0


1 2
4
| 1 0 0
→  0 −4 −12 | −4 0 4 
0 0
4
| 1 1 −3


1 2
4
| 1 0 0
→  0 −4 −12 | −4 0 4 
0 0
4
| 1 1 −3


2 4 0 | 0 −2 6
→  0 −4 0 | −1 3 −5 
0 0 4 | 1
1 −3


2 0 0 | −1 1 1
→  0 −4 0 | −1 3 −5 
0 0 4 | 1 1 −3
so

A−1
=
−2
1
1
4
1
3
2
−3
1

2
5 
−3
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