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Simon Fraser University
Math 232 Final Exam Solution
1. T F T T F F F F T T T T
a.
b.
1 2 3 3 1
2 3 3 3 2
3
r r r r r
r r r r r
↔ ← −
↔ ← −
Then if
1
2
3
x
x x
x
we have that
x s
for s ∈ .
a.
1 2
3 3 1
2
det det 7 r r
r r r
B A
↔
← +
b. det B ≠ 0 so B is invertible.
a.
det
det
i
i
A b
x
A
b.
A
and
b
, so we have
det det 1 0 2 0 6 4 0 2 4 4
A b
det A = 0 + 6 − 1 − 0 + 3 − 4 = 4 and
3
det 4
det 4
A b
x
A
2
1, , t t is a basis for 2
P.
0
p = 1
0
1 0
0 0
t p t
p t p t t t t
p p
2 2 2 2
1 0 2 2
2 1 0
1 1 0 0
(^2 2 2 )
3 3
t p t p t t t
p t p p t t
p p p p t t
t t t t t
So we have that { }
2 2
3
1, , t t − is an orthogonal basis for 2
P.
1 1 1 2 2 1 1 2 2
(^2 2 1 3 3 )
3 3 1
2
(^2 )
2
r r r r r r r
r r r (^) r r r
r r r
A
←− ← − ←
← − (^) ← +
← +
From the reduced echelon form of A we see that
C
^
B ,
N
^
B , and
R
^
B are bases for Col A , Nul A , and Row A , respectively.
a. Let [ ]
x
A
y
B
, then
x y
This gives a system of linear equations whose augmented matrix is
2 2 1 3 3 2
3 3 1
4 4 1
r r r r r r
r r r
r r r
← + ← +
← +
← −
Then
x s
(for any s ∈ ) are the least-squares solutions of Ax = b
c.
s
Ax
s
Then
2 2 2
b Ax
is the least-squares error.
a. A transformation T : V → W is said to be linear if
T ( u + v ) = T ( u ) + T ( v )
T ( cu ) = cT ( u )
for all u v , ∈ V
and all scalars c.
b. Let 2
p q , ∈ P , then
T (^) ( p t ( (^) ) + q t ( (^) )) = (^) ( 1 − t (^) ) ⋅ (^) ( p t ( (^) ) + q t ( (^) )) = (^) ( 1 − t (^) ) ⋅ p t ( (^) ) + (^) ( 1 − t q t ) ( (^) ) = T (^) ( p t ( (^) )) + T (^) ( q t ( ))
T (^) ( c ⋅ p t ( (^) )) = (^) ( 1 − t (^) ) ⋅ (^) ( c ⋅ p t ( (^) )) = c ⋅ (^) ( 1 − t (^) ) ⋅ p t ( (^) ) = c T ⋅ (^) ( p t ( ))
And so T is linear.
c. ( ) ( ) ( ) [ ]
2 2 2 3
M T T t T t t t t t t
= = − ^ − ^ ^ − =
^ ^ ^
C C C (^) C C C
is the matrix of T relative to B and C.
a. A is orthogonally diagonalisable if there is an orthogonal matrix P and a diagonal matrix D such
that
T
A = PDP.
b. A is symmetric.
c.
(^2 )
det det 1 5 1 5 1 5
A
f A I
λ
λ λ λ λ λ
λ
λ λ
The eigen-values of A are –4 and 6.
λ = − 4 :
A I
so that 4
v −
is an eigen-vector of A.
λ = 6 :
A I
−^ −
so that 6
v
is an eigen-vector of A.
We normalize 4
v −
and 6
v
as the columns of P need to be unit vectors:
1
2
4 4 1
4 2
u v
v
−
− −
−
1
2
6 6 1
(^6 )
u v
v
Then let ( )
1 1
2 2
4 6 1 1
2 2
P u u
−
−
and
D
, then
T
A = PDP.
d.
1 1 1 1 1 1
(^2 2 2 2 2 ) 1 1 1 1
4 4 6 6 1 2 2 1 2 2 1 1 1 1
2 2 2 2 2 2
T T
A u u u u
− −
−
− − −
a. S is linearly independent if whenever we choose scalars 1
p
c … c such that 1 1
p p
c v + + c v =
we must have that 1
p
c = … = c =.
b. Let T : V → W be a linear transformation of vector spaces V and W , and suppose that
p
v v ⊆ V
…. If the set ( )
p
T v T v
… is linearly independent, prove that the set
p
v v
… is also linearly independent.
Suppose that { }
1
p
v v
… is linearly dependent. Then there are scalars 1
p
c … c , not all zero
such that 1 1
p p
c v + + c v =
…. Then
1 1 1 1
p p p p
= T = T c v + + c v = c T v + + c T v
and we have constructed a non-trivial linear combination of ( )
1
p
T v T v
… equaling zero.
