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Determine value of '$a$' for which the system is inconsistent and has infinitely many solutions.

0

$begingroup$

Consider the matrix $A$, to be equal to:
begin{bmatrix}1&2&1\-1&4&3\2&-2&aend{bmatrix}

Then we can rewrite this as:
begin{bmatrix}1&2&1\0&6&4\2&-2&aend{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\2&-2&aend{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\0&-6&a-2end{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\0&0&a+2end{bmatrix}

Now consider the system $Ax=0$. If $mathbf{a = -2}$, then $x_3$ is a free variable, because a+2 turns to zero. The solution becomes:
$$x_1 = (1/3)x_3$$
$$x_2 = -(2/3)x_3$$
$$x_3 = free$$
If $mathbf{a}$ does not equal 2 , then the system seems to have trivial solutions every time, because you divide $a+2$ by itself, which becomes $1$, regardless of the value of $a$. The solutions become:
$$x_1 = 0$$
$$x_2 = 0$$
$$x_3 = 0$$

But then how can you determine when the system $Ax=0$ is inconsistent or has infinitely many solutions? What mistake have I made? Could someone provide me your own method if my method seems to be wrong?

linear-algebra matrix-equations

share|cite|improve this question

edited Mar 24 at 13:23

Later

634

asked Mar 24 at 11:54

StallmpStallmp

21219

$endgroup$

add a comment | 

0

$begingroup$

Consider the matrix $A$, to be equal to:
begin{bmatrix}1&2&1\-1&4&3\2&-2&aend{bmatrix}

Then we can rewrite this as:
begin{bmatrix}1&2&1\0&6&4\2&-2&aend{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\2&-2&aend{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\0&-6&a-2end{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\0&0&a+2end{bmatrix}

Now consider the system $Ax=0$. If $mathbf{a = -2}$, then $x_3$ is a free variable, because a+2 turns to zero. The solution becomes:
$$x_1 = (1/3)x_3$$
$$x_2 = -(2/3)x_3$$
$$x_3 = free$$
If $mathbf{a}$ does not equal 2 , then the system seems to have trivial solutions every time, because you divide $a+2$ by itself, which becomes $1$, regardless of the value of $a$. The solutions become:
$$x_1 = 0$$
$$x_2 = 0$$
$$x_3 = 0$$

But then how can you determine when the system $Ax=0$ is inconsistent or has infinitely many solutions? What mistake have I made? Could someone provide me your own method if my method seems to be wrong?

linear-algebra matrix-equations

share|cite|improve this question

edited Mar 24 at 13:23

Later

634

asked Mar 24 at 11:54

StallmpStallmp

21219

$endgroup$

add a comment | 

$begingroup$

Consider the matrix $A$, to be equal to:
begin{bmatrix}1&2&1\-1&4&3\2&-2&aend{bmatrix}

Then we can rewrite this as:
begin{bmatrix}1&2&1\0&6&4\2&-2&aend{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\2&-2&aend{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\0&-6&a-2end{bmatrix}
begin{bmatrix}1&2&1\0&1&2/3\0&0&a+2end{bmatrix}

Now consider the system $Ax=0$. If $mathbf{a = -2}$, then $x_3$ is a free variable, because a+2 turns to zero. The solution becomes:
$$x_1 = (1/3)x_3$$
$$x_2 = -(2/3)x_3$$
$$x_3 = free$$
If $mathbf{a}$ does not equal 2 , then the system seems to have trivial solutions every time, because you divide $a+2$ by itself, which becomes $1$, regardless of the value of $a$. The solutions become:
$$x_1 = 0$$
$$x_2 = 0$$
$$x_3 = 0$$

But then how can you determine when the system $Ax=0$ is inconsistent or has infinitely many solutions? What mistake have I made? Could someone provide me your own method if my method seems to be wrong?

linear-algebra matrix-equations

share|cite|improve this question

edited Mar 24 at 13:23

Later

634

asked Mar 24 at 11:54

StallmpStallmp

21219

$endgroup$

share|cite|improve this question

edited Mar 24 at 13:23

Later

634

asked Mar 24 at 11:54

StallmpStallmp

21219

share|cite|improve this question

edited Mar 24 at 13:23

Later

634

asked Mar 24 at 11:54

StallmpStallmp

21219

edited Mar 24 at 13:23

Later

634

edited Mar 24 at 13:23

Later

634

asked Mar 24 at 11:54

StallmpStallmp

21219

asked Mar 24 at 11:54

StallmpStallmp

21219

2 Answers
2

active

oldest

votes

1

$begingroup$

We are talking of the homogeneous system of equations, right?

The solution for $a=-2$ is correct. As you have said, for $aneq -2$, the only solution is $x_1=x_2=x_3=0$.

The dimension of the solution space is thus $0$ if $a=-2$ and $1$ otherwise.

share|cite|improve this answer

edited Mar 24 at 11:59

answered Mar 24 at 11:58

st.mathst.math

1,268115

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  OP is talking about infinitely many solutions, not infinitely many independent solutions
  $endgroup$
  – Kavi Rama Murthy
  Mar 24 at 11:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp There is always at least one solution for homogeneous systems of equations (the trivial solution $0$); there are infinitely many solutions for $a=-2$.
  $endgroup$
  – st.math
  Mar 24 at 12:02
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp Yes, there is always a solution.
  $endgroup$
  – st.math
  Mar 24 at 12:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp Your calculations are correct.
  $endgroup$
  – st.math
  Mar 24 at 12:06
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp: When $a=-2$, as you wrote, you have as solutions $S = { (v/3, -2v/3, v): v in mathbb{R} }$.
  $endgroup$
  – Ertxiem
  Mar 24 at 12:07
  
  
  

  
  
  
  

 | 
show 4 more comments

1

$begingroup$

$Ax=0$ has a unique solution if $det(A)neq 0$ and it has infintely many solutions if $det(A)= 0$ (any multiple of a solution is also a solution). In this case $det(A)=0$ iff $a =-2$.

share|cite|improve this answer

edited Mar 24 at 12:33

Bernard

124k742117

answered Mar 24 at 11:57

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright I see, thank you. Would this imply that inconsistency is not possible with systems Ax = 0 ?
  $endgroup$
  – Stallmp
  Mar 24 at 12:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the zero vector is always a solution, so there is no question of inconsistency.
  $endgroup$
  – Kavi Rama Murthy
  Mar 24 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright thanks a lot!
  $endgroup$
  – Stallmp
  Mar 24 at 12:09
  

  
  
  
  

add a comment | 

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1

$begingroup$

We are talking of the homogeneous system of equations, right?

The solution for $a=-2$ is correct. As you have said, for $aneq -2$, the only solution is $x_1=x_2=x_3=0$.

The dimension of the solution space is thus $0$ if $a=-2$ and $1$ otherwise.

share|cite|improve this answer

edited Mar 24 at 11:59

answered Mar 24 at 11:58

st.mathst.math

1,268115

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  OP is talking about infinitely many solutions, not infinitely many independent solutions
  $endgroup$
  – Kavi Rama Murthy
  Mar 24 at 11:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp There is always at least one solution for homogeneous systems of equations (the trivial solution $0$); there are infinitely many solutions for $a=-2$.
  $endgroup$
  – st.math
  Mar 24 at 12:02
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp Yes, there is always a solution.
  $endgroup$
  – st.math
  Mar 24 at 12:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp Your calculations are correct.
  $endgroup$
  – st.math
  Mar 24 at 12:06
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp: When $a=-2$, as you wrote, you have as solutions $S = { (v/3, -2v/3, v): v in mathbb{R} }$.
  $endgroup$
  – Ertxiem
  Mar 24 at 12:07
  
  
  

  
  
  
  

 | 
show 4 more comments

1

$begingroup$

We are talking of the homogeneous system of equations, right?

The solution for $a=-2$ is correct. As you have said, for $aneq -2$, the only solution is $x_1=x_2=x_3=0$.

The dimension of the solution space is thus $0$ if $a=-2$ and $1$ otherwise.

share|cite|improve this answer

edited Mar 24 at 11:59

answered Mar 24 at 11:58

st.mathst.math

1,268115

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  OP is talking about infinitely many solutions, not infinitely many independent solutions
  $endgroup$
  – Kavi Rama Murthy
  Mar 24 at 11:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp There is always at least one solution for homogeneous systems of equations (the trivial solution $0$); there are infinitely many solutions for $a=-2$.
  $endgroup$
  – st.math
  Mar 24 at 12:02
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp Yes, there is always a solution.
  $endgroup$
  – st.math
  Mar 24 at 12:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp Your calculations are correct.
  $endgroup$
  – st.math
  Mar 24 at 12:06
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Stallmp: When $a=-2$, as you wrote, you have as solutions $S = { (v/3, -2v/3, v): v in mathbb{R} }$.
  $endgroup$
  – Ertxiem
  Mar 24 at 12:07
  
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

We are talking of the homogeneous system of equations, right?

The solution for $a=-2$ is correct. As you have said, for $aneq -2$, the only solution is $x_1=x_2=x_3=0$.

The dimension of the solution space is thus $0$ if $a=-2$ and $1$ otherwise.

share|cite|improve this answer

edited Mar 24 at 11:59

answered Mar 24 at 11:58

st.mathst.math

1,268115

$endgroup$

share|cite|improve this answer

edited Mar 24 at 11:59

answered Mar 24 at 11:58

st.mathst.math

1,268115

answered Mar 24 at 11:58

st.mathst.math

1,268115

answered Mar 24 at 11:58

st.mathst.math

1,268115

1

$begingroup$

$Ax=0$ has a unique solution if $det(A)neq 0$ and it has infintely many solutions if $det(A)= 0$ (any multiple of a solution is also a solution). In this case $det(A)=0$ iff $a =-2$.

share|cite|improve this answer

edited Mar 24 at 12:33

Bernard

124k742117

answered Mar 24 at 11:57

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright I see, thank you. Would this imply that inconsistency is not possible with systems Ax = 0 ?
  $endgroup$
  – Stallmp
  Mar 24 at 12:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the zero vector is always a solution, so there is no question of inconsistency.
  $endgroup$
  – Kavi Rama Murthy
  Mar 24 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright thanks a lot!
  $endgroup$
  – Stallmp
  Mar 24 at 12:09
  

  
  
  
  

add a comment | 

1

$begingroup$

$Ax=0$ has a unique solution if $det(A)neq 0$ and it has infintely many solutions if $det(A)= 0$ (any multiple of a solution is also a solution). In this case $det(A)=0$ iff $a =-2$.

share|cite|improve this answer

edited Mar 24 at 12:33

Bernard

124k742117

answered Mar 24 at 11:57

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright I see, thank you. Would this imply that inconsistency is not possible with systems Ax = 0 ?
  $endgroup$
  – Stallmp
  Mar 24 at 12:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the zero vector is always a solution, so there is no question of inconsistency.
  $endgroup$
  – Kavi Rama Murthy
  Mar 24 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright thanks a lot!
  $endgroup$
  – Stallmp
  Mar 24 at 12:09
  

  
  
  
  

add a comment | 

$begingroup$

$Ax=0$ has a unique solution if $det(A)neq 0$ and it has infintely many solutions if $det(A)= 0$ (any multiple of a solution is also a solution). In this case $det(A)=0$ iff $a =-2$.

share|cite|improve this answer

edited Mar 24 at 12:33

Bernard

124k742117

answered Mar 24 at 11:57

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

share|cite|improve this answer

edited Mar 24 at 12:33

Bernard

124k742117

answered Mar 24 at 11:57

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

edited Mar 24 at 12:33

Bernard

124k742117

edited Mar 24 at 12:33

Bernard

124k742117

answered Mar 24 at 11:57

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

answered Mar 24 at 11:57

Kavi Rama MurthyKavi Rama Murthy

75.3k53270