Solve the homogeneous system of linear equations

[frctl]

x₁ − 2x₂ − 8x₃ = 0
3x₁ + 2x₂ + 0 = 0

augmented matrix:
[ 1 -2 -8 0 ]
2 2 0 0
-2R₁ + R₂
[ 1 -2 -8 0 ]
0 6 16 0
-1/6R₁ + R₂
[ 1 -2 -8 0 ]
0 1 8/3 0

Is this correct?

 

[Dr.Peterson]

Did you forget to proofread before submitting this? Check the numbers in your augmented matrix; one is wrong (unless you copied the problem incorrectly).

 

[frctl]

You're right, I have made my corrections:

[ 1 -2 -8 0 ]
3 2 0 0
-3R₁ + R₂
[ 1 -2 -8 0 ]
0 8 24 0
1/8R₂
[ 1 -2 -8 0 ]
0 1 3 0

 

[frctl]

I have arrived at my conclusions, thank you.

 

[HallsofIvy]

And what were those conclusions? What are \(\displaystyle x_1\), \(\displaystyle x_2\), and \(\displaystyle x_3\)?

Personally, I wouldn't use matrices at all. Perhaps I am just too "simple" for that! I would observe that adding the two equations immediately eliminates \(\displaystyle x_2\) leaving \(\displaystyle 4x_1- 8x_3= 0\) so that \(\displaystyle x_1= 2x_3\). And, of course, \(\displaystyle x_2= \frac{3}{2}x_1= \frac{3}{2}(2x_3)= 3x_3\). Taking \(\displaystyle x_3= a\), \(\displaystyle (x_1, x_2, x_3)= (2a, 3a, a)\).

 

[Deleted member 4993]

x₁ − 2x₂ − 8x₃ = 0
3x₁ + 2x₂ + 0 = 0

augmented matrix:
[ 1 -2 -8 0 ]
2 2 0 0
-2R₁ + R₂
[ 1 -2 -8 0 ]
0 6 16 0
-1/6R₁ + R₂
[ 1 -2 -8 0 ]
0 1 8/3 0

Is this correct?

No - the second row of your augmented matrix is incorrect.

For this homogeneous equation, only the trivial equation:

x₁ = x₂ = x₃ = 0

is possible.

 

[Steven G]

And what were those conclusions? What are \(\displaystyle x_1\), \(\displaystyle x_2\), and \(\displaystyle x_3\)?

Personally, I wouldn't use matrices at all. Perhaps I am just too "simple" for that! I would observe that adding the two equations immediately eliminates \(\displaystyle x_2\) leaving \(\displaystyle 4x_1- 8x_3= 0\) so that \(\displaystyle x_1= 2x_3\). And, of course, \(\displaystyle x_2= \frac{3}{2}x_1= \frac{3}{2}(2x_3)= 3x_3\). Taking \(\displaystyle x_3= a\), \(\displaystyle (x_1, x_2, x_3)= (2a, 3a, a)\).

\(\displaystyle x_2= -\frac{3}{2}x_1= -\frac{3}{2}(2x_3)= -3x_3\)

 

[Steven G]

No - the second row of your augmented matrix is incorrect.

For this homogeneous equation, only the trivial equation:

x₁ = x₂ = x₃ = 0

is possible.

Subhotosh, Prof Halls did get a non trivial solution (one you fix the minor error)

 

[HallsofIvy]

One of these days, I really need to learn arithmetic!