101st_tut1sol - MATH1111/2010-11/Tutorial I Solution 1...

MATH1111/2010-11/Tutorial I Solution
1
Tutorial I Suggested Solution
1. Consider the system
8
<
:
x
1
± x
2
+ x
3
± x
4
+ x
5
= 1
x
2
± x
4
± x
5
= ± 1
x
3
± 2 x
4
+ 3 x
5
= 2 :
Determine with explanation whether the following is true.
(a) There are two free variables.
(b) The free variables must be x
4
and x
5
.
Ans .
(a) The augmented matrix is already in row echelon form,
0
@
1 ± 1 1 ± 1 1
0 1 0 ± 1 ± 1
0 0 1 ± 2 3
1
± 1
2
1
A
:
Hence there are 2 free variables. In fact, take x
4
= ± and x
5
= ² , then
x
3
= 2+2 ± ± 3 ²; x
2
= ± 1+ ± + ²; x
1
= 1 + ( ± 1 + ± + ² ) ± (2 + 2 ± ± 3 ² ) + ± ± ²
= 3 ² ± 2 :
The solution set is
8
>
>
>
>
<
>
>
>
>
:
0
B
B
B
B
@
3 ² ± 2
± + ² ± 1
2 ± ± 3 ² + 2
±
²
1
C
C
C
C
A
: ±;² 2 R
9
>
>
>
>
=
>
>
>
>
;
=
8
>
>
>
>
<
>
>
>
>
:
±
0
B
B
B
B
@
0
1
2
1
0
1
C
C
C
C
A
+ ²
0
B
B
B
B
@
3
1
± 3
0
1
1
C
C
C
C
A
+
0
B
B
B
B
@
± 2
± 1
2
0
0
1
C
C
C
C
A
: ±;² 2 R
9
>
>
>
>
=
>
>
>
>
;
:
(b) No. We rewrite the system into
8
<
:
x
1
± x
2
± x
4
+ x
3
+ x
5
= 1
x
2
± x
4
± x
5
= ± 1
± 2 x
4
+ x
3
+ 3 x
5
= 2 ;
whose augmented matrix is reduced to
0
@
1 ± 1 ± 1 1 1
0 1 ± 1 0 ± 1
0 0 1 ±
1
2
±
3
2
1
± 1
± 1
1
A
:
Now, x
3
and x
5
are free variables.

MATH1111/2010-11/Tutorial I Solution
2
2. This question concerns how to prove or disprove a statement. We start with "common sense".
For statments (a) and (b) below, think about what you need to do if you want to prove it. Also,
think about how you can disprove it.
(a) All apples are sweet.