which one of the following conditions must p. q and r satisfy so that the following system of linear simultaneous equations has atleast one solution, such that (p+q+r) is not equal to zero?
x+2y-3z= p
2x+6y-11z=q
x-2y+7z=r
(1) 5p-2q-r=0
(2) 5p+2q+r=0
(3) 5p+2q-r=0
(4) 5p-2q+r=0
2006-10-20 01:42:32 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
matrice is plural already.
I am so ashamed I forgot these, but it's fun to restudy
2006-10-20 01:51:56 · answer #1 · answered by rhnegatif 2 · 0⤊ 1⤋
(1) 5p - 2q - r = 0
First you have to row reduce the matrix into triangular form to "solve" for the final row.
The row operations are:
r2 = r2 - 2r1
r3 = r3 - r1
r3 = r3 + 2r2
We then see that the last row is all zeros except for the last entry which is now
r - p + 2(q - 2p)
For this system to have a solution r - p + 2(q - 2p) must be equal to zero as well. So:
r - p + 2(q - 2p) = 0
r - p + 2q - 4p = 0
-5p + 2q + r = 0
Which is the same as 5p - 2q - r = 0 or solution (1).
2006-10-20 09:08:37 · answer #2 · answered by Leah H 2 · 0⤊ 1⤋
if you reduce your system you will find that it is
(1)5p-2q-r=0
2006-10-20 09:59:13 · answer #3 · answered by locuaz 7 · 0⤊ 0⤋