1
Simon Fraser University
Math 232 Midterm 2 Solutions
1. F T F F F T
2.
a.
( ) ( )
1 2 1 2
12 12
0 2
1 det 1 det 4
2 4
C A
+ +
= − = − =
( ) ( )
2 3 2 3
23 23
3 6
1 det 1 det 3
2 3
C A
+ +
= − = − =
( ) ( )
3 1 3 1
31 31
6 7
1 det 1 det 2
2 2
C A
+ +
= − = − = −
b.
3
11 21 31 2
1
12 22 32
3
13 23 33 2
2 3 2 1 1
1 1
4 2 6 2 1 3
det 2
4 3 6 2 3
C C C
A C C C
AC C C
−
−
− − −
= = − − = − −
− −
3.
a.
3 3 1 3 3 2
2
1 3 3 4 1 3 3 4
0 1 2 5 0 1 2 5
det det det 0
0 1 2 5 0 0 0 0
3 7 5 2 3 7 5 2
r r r r r r
A
← − ← +
− −
− −
= = =
− −
− − − − − −
b.
det 0
A
=
, so A is not invertible.
4.
a.
Because
1
P
is isomorphic to
2
through the standard co-ordinate map (that is, the map
that sends the standard basis vectors of
1
P
to the standard basis vectors of
2
),
P P
′ ′
← ←
=
C B C B
where
[ ] [ ]
{ }
1 1
1 , 1 ,
1 1
t t
′= + − =
−
E E
B
and
[ ] [ ]
{ }
2 1
2 , 1 2 ,
1 2
t t
′= + + =
E E
C
, and where
{
}
1,
t
=
E
is the standard basis for
1
P
.
To solve this equivalent problem:
( )
( )
1
3
1
3
1
2 1 1 1 1 2 1 1 1 0
~ ~ 1
1 2 1 1 0 3 1 3 0 1
C B I P
′ ′
←
−
= =
−
− − −
C B
And so
1
3
1
3
1
1
P P
′ ′
← ←
= =
−
C B C B
.
