# Thread: Linear Algebra Conditions

1. ## Linear Algebra Conditions

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

2. ## Re: Linear Algebra Conditions

In order to have a unique solution for x and y the equations must be linearly independent, which means the determinant of the coefficients must not equal 0:

$\left | \begin {matrix} 1 & b\\ 2a & 2 \end{matrix} \right | \ne 0$

Hence: $1 \times 2 - b \times 2a \ne 0$

which means $ab \ne 1$

3. ## Re: Linear Algebra Conditions

If you haven't yet covered "determinant", a more basic way is to try to solve the system of equations. The given equations are x+ by= -1 and 2ax+ 2y= 5. From the first equation , x= -1- by. Replace x by that in the second equation: 2a(-1- by)+ 2y= -2a- 2aby+ 2y= (2- 2ab)y- 2a= 5.
So (2- 2ab)y= 5. We can solve for the unique value y= 5/(2- 2ab), and then for a unique value of x if and only if that denominator, 2- 2ab= 2(1- ab) is not 0. That is the same as saying ab is not 1.