Question 89229
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Determine the value(s) of for which the following system of equations has
nontrivial solutions. 
(k+4)x + (k-2)y = 0
4x + (k-3)y = 0 
Please explain how you get the answer so I can do the others.

"Non-trivial" means other than (x,y) = (0,0)

{{{(matrix(2,2,k+4,k-2,4,k-3))}}}{{{(matrix(2,1,x,y))}}} = {{{(matrix(2,1,0,0))}}}

If the coefficient matrix {{{ (matrix(2,2,k+4,k-2,4,k-3)) }}}
has an inverse, we could left-multiply both sides by that inverse and
there could only be the trivial solution.  That's because that inverse
times {{{(matrix(2,1,0,0))}}} on the right could only give {{{(matrix(2,1,0,0))}}} on the right

So there could only be a nontrivial solution if the coefficient matrix
is singular (has no inverse).  This only occurs when its
determinant is 0. (That is, when Cramer's rule fails).

Thus we set its determinant = 0

{{{abs(matrix(2,2,k+4,k-2,4,k-3))}}} = 0

 (k+4)(k-3) - 4(k-2) = 0

Multiply that out, solve it and get two solutions

k = -1 and k = 4

Using k = -1,

(-1+4)x + (-1-2)y = 0
4x + (-1-3)y = 0

3x - 3y = 0
4x - 4y = 0

Both those are equivalent to y = x, so when k = -1,
then there are infinitely many solutions, and they
all have the form (c,c) where c is any number.

Using k = 4,

(4+4)x + (4-2)y = 0
4x + (4-3)y = 0

8x - 2y = 0
4x -  y = 0

Both those are equivalent to y = 4x, so when k = 4,
then there are infinitely many solutions, and they
all have the form (c,4c) where c is any number.

Edwin</pre>