Help me please with system.
It turned out this way: Multiply system 2
4x + by = 4
2bx + 2y = 4
The system has no solutions if:
4x + by ≠ 2bx + 2y
x (4 - 2b) ≠ y (2 - b)
2x (2 - b) ≠ y (2 - b)
For any b ≠ 2, the system will have only one solution.
And when b = 2 - infinite number of solutions:
4x + 2y = 4
2x + y = 2
Does it mean that for any b there will be such that the system has no solution?
It seems to me that when b = -2 will not solutions ....
On
You have the following system with $b\not = 0$: $$4x+by=4\Rightarrow y=-\frac{4}{b}x+\frac{4}{b}$$ $$2bx+2y=4\Rightarrow y=-bx+2 $$ These equations are the equations of two straight lines with slopes $λ_1=-\frac{4}{b}$ and $λ_2=-b$. These two straight lines have no solution when they do not intersect, meaning that they are parallel. Hence must:
$$λ_1=λ_2\Rightarrow -\frac{4}{b}=-b\Rightarrow b^2=4\Rightarrow b=-2 \ or \ b=2$$
When $b=2$ the two equations have infinite solutions, hence it's rejected. So $b=-2$.
If $b=0$ then $x=2 \ and \ y=2$.
So after all the only solution to the problem is $b=-2$
Yes it is correct indeed the system is
and for $b=-2$
We can solve the problem in a more sistematic way by the RREF on the augmented matrix, that is
$$\left[ {\begin{array}{rr|r} 4&b&4\\ b&1&2\\ \end{array} } \right]\to \left[ {\begin{array}{rr|r} 4b&b^2&4b\\ 4b&4&8\\ \end{array} } \right]\to \left[ {\begin{array}{rr|r} 4b&b^2&4b\\ 0&4-b^2&8-4b\\ \end{array} } \right]$$
and thus the system has no solution for
that is precisely $b=-2$.