MHB Systems of equations - further understanding

[Yankel]

Hello again,

I have a few more questions regarding systems of equations, I will collect them all here in one post since they are small.

1. The first is the following system:

x+2y-3z=a
3x-y+2z=b
x-5y+8z=c

I need to determine the relation between a,b and c for which the system has infinite solution, unique solution or no solution. I did some row operations and got:

\[\begin{pmatrix} 1 &2 &-3 &a \\ 0 &-7 &11 &b-3a \\ 0 &0 &0 &2a-b+c \end{pmatrix}\]

I conclude that when 2a-b+c=0 there is an infinite solution and when it ain't equal 0, there is no solution. A unique solution is not possible. However, Maple got the same matrix but claims that there is no solution either way...is it a computer bug or I am mistaken ?

2. A is a matrix over the R field with dimensions 3X4. The rank of A is 1. How many degrees of freedom (parameters, i.e. t,s,...) does the family of solutions of Ax=0 has ?

3. If Ax=b has infinite solution, then Ax=c has infinite solution or no solution. True or False ?

Thanks a lot !
:)

 

[Petrus]

Hello again,

I have a few more questions regarding systems of equations, I will collect them all here in one post since they are small.

1. The first is the following system:

x+2y-3z=a
3x-y+2z=b
x-5y+8z=c

I need to determine the relation between a,b and c for which the system has infinite solution, unique solution or no solution. I did some row operations and got:

\[\begin{pmatrix} 1 &2 &-3 &a \\ 0 &-7 &11 &b-3a \\ 0 &0 &0 &2a-b+c \end{pmatrix}\]

I conclude that when 2a-b+c=0 there is an infinite solution and when it ain't equal 0, there is no solution. A unique solution is not possible. However, Maple got the same matrix but claims that there is no solution either way...is it a computer bug or I am mistaken ?

2. A is a matrix over the R field with dimensions 3X4. The rank of A is 1. How many degrees of freedom (parameters, i.e. t,s,...) does the family of solutions of Ax=0 has ?

3. If Ax=b has infinite solution, then Ax=c has infinite solution or no solution. True or False ?

Thanks a lot !
:)

Hello,

1. For it to be infinity soloution you want them to be linear dependen
2. Dim ker (A) Tells you how many parameters there is,

edit: 1. Yes it looks correct for me what you Said

notice that I have not checked your progress!

Regards,
$$|\pi\rangle$$

 

[Deveno]

Umm...don't trust computers, they lie to you.

OBVIOUSLY, there is the solution (0,0,0) when a = b = c = 0. perhaps not as obviously, there are also the solutions of the form:

t(-1,11,7) for any real number t, when a = b = c = 0.

Thus given some vector (a,b,c) for which 2a - b + c = 0 (like, for example: (1,1,-1)), we can conclude we have the infinite number of solutions:

(2/7,13/7,1) + t(-1,11,7), since:

A(2/7,13/7,1) = (2/7 + 26/7 - 3, 6/7 - 13/7 + 2,2/7 - 65/7 + 8) = (1,1,-1) and

A(t(-1,11,7)) = t(A(-1,11,7)) = t(0,0,0) = (0,0,0)

So clearly Maple is wrong about the number of solutions.

For #2, the rank-nullity theorem tells you that:

rank(A) + nullity(A) = 4. See also Petrus' answer above, note that, by definition:

nullity(A) = dim(ker(A))

For #3: on these types of problems it's good to play with some simple examples.

Try using:

$A = \begin{bmatrix}1&0\\0&0 \end{bmatrix}$

$b = \begin{bmatrix}1\\0 \end{bmatrix}$

and

$c = \begin{bmatrix}2\\0 \end{bmatrix}$

or

$c = \begin{bmatrix}0\\2 \end{bmatrix}$

Now suppose the statement is false:

this means that we have a UNIQUE solution x₀ of Ax = c, but infinitely many of Ax = b.

Pick two DIFFERENT solutions of Ax = b, say x = x₁, x₂.

Since these are different solutions, x₁ - x₂ ≠ 0, so x₁ - x₂ + x₀ ≠ x₀.

Now A(x₁ - x₂ + x₀) = A(x₁) - A(x₂) + A(x₀​) = b - b + c =...?