MHB Linear Algebra Conditions: Solving for ab≠1

[Terry_Destefano]

View attachment 6404
The answers is b) ab≠1, but I have no clue how to get to that answer... Can someone help me? :D

 

[HallsofIvy]

The answers is b) ab≠1, but I have no clue how to get to that answer... Can someone help me? :D

One answer to this is that, in order that a system of linear equations have a unique solution is that the determinant of coefficient, here $$\left|\begin{array}{cc}1 & b \\ 2a & 2 \end{array}\right|= 2- 2ab$$ be non-zero. (This is the answer you got to your post of this question on another board.)

More fundamentally, what happens if you try to solve the equations? To solve x+ by= -1, 2ax+ 2y= 5, we might start by multiplying the first equation by 2, 2x+ 2by= -2. Multiply the second equation by b, 2abx+ 2by= 5b. Now, since both equations have "2by", subtracting one equation from the other (2- 2ab)x= -2- 5b. Finally, divide the both sides by 2- 2ab: x= (-2- 5b)/(2- 2ab). Given that value for x, we could put it into either of the two original equations to solve for y.

Under what conditions can we not do that? Every step above is easy to do for all values of a and b except "divide both sides by 2- 2ab". We cannot do that if 2- 2ab= 0 because we cannot divide by 0. There is a unique solution for all a or b except a and b so that 2- 2ab= 2(1- ab)= 0. That is the same as ab= 1. The given system of equations has a unique solution if and only if ab is not equal to 1.

 

[Terry_Destefano]

Thanks HallsofIvy!
I put my question at many forums and even payed for homework help but you gave me the most detailed response.

 

[MarkFL]

Thanks HallsofIvy!
I put my question at many forums and even payed for homework help but you gave me the most detailed response.

There's no need to link to the commercial site you used. Thanks! :D