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- Given the following homogeneous system Ax = 0 :
x 1
x 2
x 3
x 4
x 5
[4] (a) Write the solution to the system in parametric vector form.
[1] (b) Write the zero vector in R
4 as a nontrivial linear combination of the columns of A.
[4] 2. Use techniques of linear algebra to find a polynomial p(x) = a 0 + a 1 x + a 2 x
2 such that p(2) = 0,
p(−2) = 32 and p
′ (1) = −7.
[3] 3. Let v 1 =
, v 2 =
k
, v 3 =
k
0
2 k + 3
(^) and let S = {v 1 ,^ v 2 ,^ v 3 }. For what value(s) of^ k^ is:
(a) Span(S) all of R
3 ?
(b) Span(S) a plane in R
3 ?
(c) Span(S) a line in R
3 ?
- Let T 1 : R
2 → R
2 be a linear transformation defined by T 1
([
x
y
])
[
−x + 2y
2 x − 3 y
]
[1] (a) Find the standard matrix for T 1.
[3] (b) If L is the line
[
]
[
k
]
, then for what value(s) of k, will T 1 (L) be a horizontal line in R
2 ?
[3] (c) Now suppose that the composition T 1 ◦ T 2 is also a linear transformation whose standard matrix
is
[
]
i. Identify the domain and codomain of T 2.
ii. Find the standard matrix for T 2.
[3] 5. Suppose that the set {u, v, w} is linearly independent, and that x = 2u + 3w and y = v + 2w. Prove
that the set {u, x, y} is linearly independent.
[3] 6. Let A =
. Find an LU factorization of A, where L is unit lower triangular and U is upper
triangular.
[4] 7. Let A =
a d g
b e h
c f k
(^) and B =
a + 2b + 4c d + 2e + 4f g + 2h + 4k
3 a + 4b + 7c 3 d + 4e + 7f 3 g + 4h + 7k
5 a + 7b + 8c 5 d + 7e + 8f 5 g + 7h + 8k
(a) Find a matrix C such that B = CA.
(b) Find the value of λ such that det B = λ det A for all possible choices of A.
[4] 8. Let A be a 3 × 3 matrix and let det A = −2.
(a) Find det (A
T A
2 (− 2 A)
− 1 ).
(b) Find det (adj (2A)).
[6] 9. (a) Find matrices W , X, Y and Z such that
[
O A
B O
] [
W X
Y Z
]
[
I O
O I
]
(where A and B are
invertible matrices).
(b) Use the above result to find C
− 1 , where C =
[3] 10. Use Cramer’s Rule to solve the system:
7 x − 9 y = 11
4 x + 5y = − 2
[3] 11. Simplify the matrix expression (B(B + I)
− 1 )
− 1 − B
− 1
[6] 12. Given A =
∼ R =
(a) Row A is a subspace of R
n for what value of n?
(b) Without calculation, give a basis for Row A.
(c) Col A is a subspace of R
m for what value of m?
(d) Without calculation, give a basis for Col A.
(e) What is rank A
T ?
(f) What is dim Nul A
T ?
[4] 13. Let W =
{[
x 1
x 2
]
∈ R
2 : x 1 = 0 or x 2 = 0
(a) Is 0 in W? Justify your answer.
(b) Is W closed under scalar multiplication? Justify your answer.
(c) Is W closed under vector addition? Justify your answer.
(d) Is W a subspace of R
2 ? Explain.
[3] 14. Let A =
[
]
. Find a 2 × 2 matrix B such that AB = O but BA 6 = O (where O is the zero
matrix).
[6] 15. In question 14 you saw that there can be non-zero n × n matrices A and B such that AB = O but
BA 6 = O. Now let A and B be any two such matrices.
(a) Show that each column of B is in Nul A.
Answers
- (a) x = s
(b) Use part (a) and give non-zero values to s and/or t to generate a set of weights; for instance
s = 1 and t = 1 gives x = (1, − 1 , 1 , 1 , 0):
- p(x) = 14 − 8 x +
1 2
x
2
- (a) k 6 = − 1 , 3 (b) k = − 1 , 3 (c) no value of k
- (a)
[
]
(b)
(c) i. Domain is R
3 , codomain is R
2 ii.
[
]
- You’ll want to show that a 1 u + a 2 x + a 3 y = 0 has no non-trivial solution. Making the substitutions
for x and y and rearranging, the equation becomes (a 1 + 2a 2 )u + (3a 2 + 2a 3 )w + a 3 v = 0. Since the
set {u, v, w} is linearly independent, all of the weights in the second equation much be zero. Use
standard linear algebra techniques to show that the system of equations
a 1 + 2a 2 = 0
3 a 2 + 2a 3 = 0
a 3 = 0
has only the trivial solution.
6. A =
- (a)
(^) (b) λ = 9 (the determinant of C)
- (a) −
(b) 2
8
- (a)
[
W X
Y Z
]
[
O B
− 1
A
− 1 O
]
(b) C
− 1
- x =
, y = −
- The expression simplifies to I.
- a) n = 7 b)
c) m = 5 d) b)
e) 4 f) 1
- W contains 0 and is closed under scalar multiplication, but is not closed under vector addition and
therefore is not a subspace of R
2 .
- Multiple answers are possible. One example is
[
]
- (a) Let b 1 and b 2 be the columns of B.
AB = [Ab 1 Ab 2 ] = [0 0] = O
Since Ab 1 = 0 and Ab 2 = 0 , b 1 and b 2 are in Nul A.
(b) Proof:
(BA)
2 = (BA)(BA)
= B(AB)A (associativity)
= BOA (sinceAB = 0)
= O
(c) Suppose B were invertible. Then we could do this:
AB = O
ABB
− 1 = OB
− 1
A = O
But A 6 = O. Contradiction. Therefore B is not invertible.
- Multiple answers are possible. One example is
{[
]
[
]}
- dim V = 2. One example of a basis for V is {x − 2 , x
2 − 4 }.
- a) n b) n − m
