find the value of k so that the system of the equation
x + y + 3z = 0
4x + 3y + kz = 0
2x + y + 2z = 0
has non-trival solution
This is called a homogeneous system. It has only zeros
on the right side. The trivial solution for all
homogeneous systems is (x,y,z) = (0,0,0). But many
homogeneous systems have other, non-trivial solutions.
Make the augmented matrix:
[1 1 3 | 0]
[4 3 k | 0]
[2 1 2 | 0]
To get a 0 where the 4 is,
multiply row 1 temporarily
by -4 and add it to 1 times
row 2, then restore row 1:
-4[1 1 3 | 0]
1[4 3 k | 0]
[2 1 2 | 0]
[1 1 3 | 0]
[0 -1 -12+k | 0]
[2 1 2 | 0]
To get a 0 where the 2 on
the bottom left is, multiply
row 1 temporarily by -2 and
add it to 1 times row 3, then
restore row 1:
-2[1 1 3 | 0]
[0 -1 -12+k | 0]
1[2 1 2 | 0]
[1 1 3 | 0]
[0 -1 -12+k | 0]
[0 -1 -4 | 0]
To get a 0 where the -1 on
the 3rd row is, multiply row
2 temporarily by -1 and add
it to 1 times row 3, then
restore row 2:
[1 1 3 | 0]
-1[0 -1 -12+k | 0]
1[0 -1 -4 | 0]
[1 1 3 | 0]
[0 -1 -12+k | 0]
[0 0 8-k | 0]
Multiply row 2 by -1
[1 1 3 | 0]
[0 1 12-k | 0]
[0 0 8-k | 0]
The system will have
a non-trivial solution
if the bottom row has
only zeros, hence 8-k
must = 0. So
8 - k = 0
k = 8
That's the answer.
------------------------
However, if you want to
find the non-trivial
solutions, then substitute
k = 8 in the matrix
[1 1 3 | 0]
[0 1 12-k | 0]
[0 0 8-k | 0]
which becomes
[1 1 3 | 0]
[0 1 12-(8) | 0]
[0 0 8-(8) | 0]
[1 1 3 | 0]
[0 1 4 | 0]
[0 0 0 | 0]
Rewriting as a system of
equations:
1x + 1y + 3z = 0
0x + 1y + 4z = 0
0x + 0y + 0z = 0
or to show triangular form:
x + y + 3z = 0
y + 4z = 0
0z = 0
The 3rd equation tells us
that z can equal any arbitrary
number, so substitute z = a
in the 2nd equation
y + 4z = 0
y + 4a = 0
y = -4a
Substitute z = a and y = -4a
into the first equation:
x + y + 3z = 0
x + (-4a) + 3a = 0
x - 4a + 3a = 0
x - a = 0
x = a
So all non-trivial solutions
can be found from:
(x, y, z) = (a, -4a, a)
by choosing arbitrary values
for a. The trivial solution
is when a = 0
(x, y, z) = (0, 0, 0)
One non-trivial solutions can be
found when a = 1
(x, y, z) = (1, -4, 1)
When a = 2, we get the non-trivial
solution
(x, y, z) = (2, -8, 2)
When a = -6.37, we get the non-trivial
solution
(x, y, z) = (-6.37, 25.48, -6.37)
etc., etc., etc.
Edwin
