If possible, find an example of actual numbers c1, c2, and c3, not all zero, so that the following is true.

Question

c1[3,3,−1]
+ c2[1,2,−3]
+ c3[−11,−13,9]
= [0,0,0]

Answer 1

I will use a, b, and c, instead of c1, c2 and c3

a(3,3,-1) + b(1,2,-3) + c(-11,-13,9) = (0,0,0)

3a + 2b - 11c = 0
3a + 2b - 13c = 0
-a - 3b + 9c = 0

subtract 2nd from 1st
2c = 0
c = 0
sub into 2nd and 3rd
3a + 2b = 0
-a - 3b = 0

triple the 3rd:
-3a - 9b = 0
3a + 2b = 0
-7b = 0
b = 0
clearly also a = 0

so it looks like the trivial case, that is
c1=0
c2=0
c3=0 is the only case that works,

so for your question, since you don't want the all zero case, no
there is no other solution

Answer 2

I am going to call c1 = a
c2 = b
c2 = c

+3a + 1b -11c = 0
+3a + 2b -13c = 0
-1a - 3b + 9c = 0

-b+2c = 0 from first 2 so b = 2 c

+3a + 2b -13c = 0
-3a - 9b +27c = 0 from second 2
--->
-7 b + 14 c = 0
so
-7(2c) = -14 c
c = anything at all

In your linear algebra using Cramer's method
You can get a finite determinant for your coefficient matrix
HOWEVER when you plug in the numerator matrix with a column of zeros, you always get 0 in the numerator.

Answer 3

well, we agree this time :)

Answer 4

and we both called the variables a, b, and c
What's that old saying, something about great minds, lol

Answer 5

EXAMLE 75@:~5@_2_3_2(_3)

Answer 6

what is 75@<5_2_3_2(_3)