Post by: asterio on July 05, 2010, 11:45:27 am
The simulaneous equations:
(n-3)x + 3y = 6
2x + (n+3)y= n
have a unique solution for n.
can someone plz tell me HOW do you find it...?
Post by: /0 on July 05, 2010, 12:04:37 pm
In a matrix,
There will be a unique solution provided the determinant of the coefficient matrix is non-zero. That gives you your values of n.
Post by: asterio on July 05, 2010, 12:18:58 pm
A: n (include) R\{0}
B: n (include) R\{-3,3}
C: n (include) R\{-15^0.5,15^0.5}
D: n (include) R\[-3,3]
E: n = 3 or n = -3
Post by: /0 on July 05, 2010, 12:22:13 pm
When that equals zero,
So the values of 'n' for which you will get a unique solution are
Post by: tcg93 on July 05, 2010, 10:27:59 pm
I know /0 is right, but when a question asks for unique solution, doesn't it mean that one set of solutions should exist?
When det = 0, there should be no solutions instead... (either because they lines don't intersect or the lines are equal)
Yes, so the answer is R\{-15^0.5,15^0.5} as it is all values when det does not equal 0