If possible, find an example of actual numbers c1, c2, and c3, not all zero, so that the following is true.
c1[3,3,−1]
+ c2[1,2,−3]
+ c3[−11,−13,9]
= [0,0,0]
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
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.
well, we agree this time :)
and we both called the variables a, b, and c
What's that old saying, something about great minds, lol
EXAMLE 75@:~5@_2_3_2(_3)
what is 75@<5_2_3_2(_3)
To find an example of actual numbers c1, c2, and c3 that satisfy the given equation, we can set up a system of equations and solve for the values of c1, c2, and c3.
Let's denote the given equation as:
c1[3, 3, -1] + c2[1, 2, -3] + c3[-11, -13, 9] = [0, 0, 0]
Expanding the equation, we get:
[3c1 + c2 - 11c3, 3c1 + 2c2 - 13c3, -c1 - 3c2 + 9c3] = [0, 0, 0]
Now, we can equate the corresponding components of the vectors:
3c1 + c2 - 11c3 = 0
3c1 + 2c2 - 13c3 = 0
-c1 - 3c2 + 9c3 = 0
To solve this system of equations, we can use various methods such as substitution or matrix operations. Here, we'll use matrix operations to find the values of c1, c2, and c3.
Writing the system of equations in matrix form, we have:
⎡ 3 1 -11 ⎤ ⎡ c1 ⎤ ⎡ 0 ⎤
⎢ 3 2 -13 ⎥ ⎢ c2 ⎥ = ⎢ 0 ⎥
⎣ -1 -3 9 ⎦ ⎣ c3 ⎦ ⎣ 0 ⎦
To solve for c1, c2, and c3, we need to find the row reduced echelon form of the augmented matrix [A | B], where A is the coefficient matrix, and B is the column matrix [0, 0, 0].
Performing row operations on the augmented matrix, we get:
⎡ 1 0 2 ⎤ ⎡ c1 ⎤ ⎡ 0 ⎤
⎢ 0 1 -1 ⎥ ⎢ c2 ⎥ = ⎢ 0 ⎥
⎣ 0 0 0 ⎦ ⎣ c3 ⎦ ⎣ 0 ⎦
From the row reduced echelon form, we can see that c1 = -2c3 and c2 = c3.
Now, we can substitute these values back into the original equation, c1[3, 3, -1] + c2[1, 2, -3] + c3[-11, -13, 9] = [0, 0, 0]:
(-2c3)[3, 3, -1] + c3[1, 2, -3] + c3[-11, -13, 9] = [0, 0, 0]
Simplifying the equation, we get:
[-6c3, -6c3, 2c3] + [c3, 2c3, -3c3] + [-11c3, -13c3, 9c3] = [0, 0, 0]
Combining like terms, we have:
[-16c3, -17c3, 8c3] = [0, 0, 0]
For the equation to hold true, the components of the vector on the left side must all be zero. This implies that -16c3 = 0, -17c3 = 0, and 8c3 = 0.
From these equations, we find that c3 = 0.
Substituting c3 = 0 back into the expressions for c1 and c2, we get c1 = -2(0) = 0 and c2 = 0.
Thus, an example of actual numbers c1, c2, and c3 that satisfy the given equation is c1 = 0, c2 = 0, and c3 = 0.