Download Linear algebra 1A - (partial) solution of ex.2

Linear algebra 1A - (partial) solution of ex.2
2) Find for what values of
t the following linear system has a unique solution,
no solution, innite number of solutions. Work over the real numbers. You dont
need to nd the explicit sets of solution


tx + 3y − z
x + 2y − z


−tx + y + 2z
=1
=2
= −1
-SolutionLet us write this system (in variable x,y,z) as a matrix and start performing
elementary row operations (as usual).
There is more than one way to proceed - we do not use here almost the
standard algorithm, but with a slight variation, in order to make the calculations
easiers. Any proccess that uses only elementary operations is alright.



2
1
1 2 −1
t 3 −1
 1 2 −1 | 2  R1 ↔ R2  t 3 −1 | 1  R2 7→ R2 + R3
→
−→
−1
−1
−t 1 2
−t 1 2




1
2
−1
2
1 2 −1
2
1
 0 4 1 | 0  R3 7→ R3 + tR1 , R2 7→ R2 /4  0

0
1
|
4
→
−1
−t 1 2
0 1 + 2t 2 − t −1 + 2t


1 2
−1
2
1

0
|
R3 7→ R3 − (1 + 2t)R2  0 1
4
7
→
0 0 4 − 32 t −1 + 2t
7
3
7
7
Now, if − t = 0, i.e. if t = , then in the last line we have (0 0 0 | −1+ ),
4
2
6
3

so there are no solutions to the system.
If
t ̸=
7
6 , then we can devide
So, for any
t ̸=
R3 by 74 − 32 t =

7−6t
4
1
, to get  0
0
2
1
0
−1
1
4
|
2
0
4(−1+2t)
1
7−6t
7
6 the system has one solution exactly. One can compute it,
if needed.
(Thus, in this case, only no solutions and one solution are possible. Generally, the third option innitely many can occur.)
5) Suppose that the linear system
{
ax + by = 0
(♯)
cx + dy = 0
numbers r and s, the
has a unique solution. Prove that for any choice of
linear system
1

.
{
ax + by = r
(♭)
cx + dy = s
has a unique solution.
Proof: The unique solution of the homogeneous system
(♯)
is
(0, 0),
and the
canonical form of the corresponding matrix is then
(
)
1 0 0
. This matrix is achieved
0 1 0
E1 , ..., EM of elementary row operations.
(
)
a b 0
by applying a list
c d 0
(
)
a b r
Now, consider the marix corresponding to (♭) :
.
c d s
Applying the same elementary row operations E1 , ..., EM to this matrix, we
)
(
1 0 r′
′ ′
get a matrix of the form
(where r , s some numbers in our eld).
0 1 s′
from
This last system has a unique solution, because the matrix on the left is the
identity matrix (matritzat ha-yekhida). (By a theorem studied.)
(
)
(
)
1 0 r′
a b r
and
(♭)
are equivalent, hence
But this last matrix
′
0 1
(♭)
also has a unique solution.
c
s
d
s
6) Given a linear system, denote by A the corresponding reduced echelon
matrix.
a) If the system is homogenous, and if A has ve leading terms, what is the
minimal number of variables that the system can possibly have?
- Minimal number of variables: ve. Because each leading term correspondes
one column exactly, and each column correspondes to one variable. (These ve
are mishtanim tluim (dependent variables)).
b) If the system is homogenous and has 5 variables and a unique solution,
what is the minimal and maximal number of leading terms in the matrix A?
- If a homogeneous system has a unique solutions, then this is the trivial
(zero) solution, and the corresponding reduces form is the identity matrix, so
A
must have ve leading terms exactly.
c) Is it possible that the leading term which appears in the fth row of A,
be located in the sixth column of A? in its fourth column?
- In the sixth - possible (if A has at least six columns). Here is an example:






1
0
0
0
0
9
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1



.


In the forth - impossible. This would mean that the rst ve leading terms
are located in the rst 4 columns; But each leading term should correspond to
exactly one column and each column corresponds to not more than one leading
term.
d) If A has ve columns, what is the minimal number of leading terms the
matrix A can have?
- zero. This can be the zero matrix, with ve (all) free variables.
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