Hey so for questions that ask you:
Solve the following system of equations and find the value of k, such that a unique solution, infinite number of solutions or no solutions are given.
Do we use the determinant of the inverse matrix? I was taught this briefly by my tutor but I forgot how to do it.. Can someone elaborate? Was it determinant equal zero means there is one unique solution or something like that?
Also, would it be better if we solved it without matrices? Ie. Actually solve the simultaneous equation by hand, and apply logic (i.e.. since there are an infinite number of solutions, gradient must be the same etc.) without using matrices?
Thanks!
If the determinant of the (square) coefficient matrix is non-zero, then there is a unique solution. If it is zero, then there could be zero or infinitely many solutions.
The usual method to solve this problem would be to find the reduced row echelon form of the augmented matrix and see how the solutions depend on k.
If you have all zeroes in a row on one side of the line and a non-zero number on the other, the system has no solutions. If you have all zeroes in a row, on either side of the line, the system has infinitely many solutions. If you have no non-zero rows then the system has a unique solution.
*ahem this is VCE. No row reducing allowed here
also 'no non-zero rows?' That one I believe IS a typo
For VCE purposes, as kinslayer has mentioned, check if the coefficient matrix determinant is zero. If it is zero, you should get a few values for your parameter. If you plug those into the original equations, you should either get identical lines (like x+y = 2 and 2x + 2y = 4) or parallel lines (x+y=2 and 2x+2y=2). Identical lines => infinitely many solutions
Parallel lines => no solution
As for the determinant of the inverse matrix, for a 2 x 2 matrix (which is all you'll need for Methods by hand) it's the product of the top left and bottom right minus the top right times bottom left.
If your matrix is
a     b
c     d
Your determinant is ad - bc
However if you're not comfortable with matrices, you CAN solve it by hand. Just make sure you're careful to not divide by any parameters. For instance, kx + ky = 0 doesn't necessarily mean x+y=0, it might just mean k=0, as an example.
You can go from x+y = 1 to kx + ky = k
but not the other way around. Why? If k happens to be zero, you will have divided by zero which you can't do.
As an example, consider kx + 3y = k-1
                                       2x + (k-1)y = 1
Double first equation, multiply second by k
2kx + 6y = 2k-2
2kx + k(k-1)y = k
What you can't do is double the first equation and divide it by k, for instance
No solutions when gradients are equal => same y coefficients
6 = k(k-1)
k=3 or k=-2
If k=3, our equations are 6x + 6k = 4
                                 and 6x + 6k = 3. Parallel lines
If k=-2, we have
-4x + 6y = -6
-4x + 6y = 2
Again parallel lines
We'd get the same line if the second equation were -4x + 6y = -6 as well