system with constant

[bluemath]

Hello,

We have :

ax + 2y + z = 1
(b-1)y + az = a
ax + 2by + bz = 1

I found that if b = 2a + 1 , then there is no solutions.

Now how to find value for a et b who give an only solution for the system ?

Same thing but with one free parameter ?

Thanks in advance

 

[Deleted member 4993]

Hello,

We have :

ax + 2y + z = 1
(b-1)y + az = a
ax + 2by + bz = 1

I found that if b = 2a + 1 , then there is no solutions.

Now how to find value for a et b who give an only solution for the system ?

Same thing but with one free parameter ?

Thanks in advance

For a unique, non-trivial solution - the determinant must be non-zero.

 

[Deleted member 4993]

Hello,

We have :

ax + 2y + z = 1
(b-1)y + az = a
ax + 2by + bz = 1

I found that if b = 2a + 1 , then there is no solutions. .... What if b=1?

Now how to find value for a et b who give an only solution for the system ?

Same thing but with one free parameter ?

Thanks in advance

.

 

[Ishuda]

Hello,

We have :

ax + 2y + z = 1
(b-1)y + az = a
ax + 2by + bz = 1

I found that if b = 2a + 1 , then there is no solutions.

Now how to find value for a et b who give an only solution for the system ?

Same thing but with one free parameter ?

Thanks in advance

They way I would tackle it would be to substitute c=b-1 and subtract row 1 from row 3 to get
\(\displaystyle \begin{pmatrix}
ax& 2y& z& =& 1\\
0& cy& az& =& a\\
0&2cy& cz& =& 0
\end{pmatrix}\)
Now subtract twice row 2 from row 3 to get
\(\displaystyle \begin{pmatrix}
ax& 2y& z& =& 1\\
0& cy& az& =& a\\
0& 0& (c-2a)z& =& -2a
\end{pmatrix}\)

So Start from the bottom up:
(1)If c=2a then a=c=0 [b=1] and z=1-2y.
(2)If c is not equal to 2a then z=-\(\displaystyle \frac{2a}{c-2a}\) and going to the second row,
__(2a) If a is zero then
______(2ai) If c is zero that is case (1)
______(2aii) If c is not zero then y is zero and go to row 1 ...
__(2b) If a is not zero ...

 

[bluemath]

Thanks a lot for all your answers.

I did a misstake, correction : if b = 2a + 1 AND a not egal to zero, then we have no solutions.

 

[Deleted member 4993]

Sorry Subhotosh...lost your red ink

Doing the rock and roll with the 3 equations leads to:
z = 2a / (2a + 1 - b) ; so 2a + 1 <> b

I was trying to point out that for b=1 there are infinite solutions.