Question

given the matrix | 1 3 1|
M = | 2 0 -1| where det(M) = 0
| 1 1 0|
the equation MX = 0 will have solutions other than the trivial solution X = 0. How do I find these other solutions?

Answer 1

\[M = \left[\begin{matrix}1 & 3 & 1 \\ 2 & 0 &-1\\ 1 & 1 & 0\end{matrix}\right]\]

Answer 2

by calculating the eigene values and vectors

det(M-LI) forms a cubic equation to solve for L, those are your values

Answer 3

-(-L-1+L-6L)
1-L 2 1 1-L 2
3 -L 1 3 -L
1 −1 -L 1 -1
(L^2-L^3+2-3)

L^2 - L^3 + 2 - 3 +L +1- L+ 6L

L^2 - L^3 + 6L= 0

L (- L^2 +L + 6) = 0 ; when L=0

L^2 -L - 6 = 0 ; when L = 3, -2

Answer 4

Im not sure what eigene values are. In the problem it also says:

"Hint: call U, V and W the three rows of M, and
observe that M X = 0 if and only if X is orthogonal to the vectors U, V and W."

so do they mean there is a way to solve it geometrically? or maybe they are hinting towards something that I havent learned yet.

Answer 5

hmm, i think im confusing some stuff :)

MX = X has to do with eigene values ....

Answer 6

1x +2y +1z = 0
3x +0y +1z = 0
1x −1y + 0z = 0

per their hint, you have a system of 3 equation in 3 unknowns

Answer 7

Well this is a special case. Apparently, if the determinate of M isn't 0, then MX = 0 will only have one trivial solution X = 0. But since it is 0, the worksheet implies that there are other solutions based on the fact that "U, V and W" will form a parallelepiped with no volume. I am not sure what they mean. Here is the worksheet, I am on Problem 3 C http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/exams/prac1a.pdf

Answer 8

Ooooooohhh, thank you that explains it.

Answer 9

I am kind of confused though on how the determinant of a square matrix finds the volume of a parallelepiped. Was the definition of a determinant made such that A•(B*C) = the determinant of a matrix with the components of the vectors a b and c? or is this derived elsewhere?

Answer 10

if we take the matrix as is, and row reduce it we get

1 0 1/3
0 1 1/3
0 0 0

this tells us a nul space; or rather, all the vectors that form MX = 0, by solving x and y in terms of z. does that make sense?

x = -z/3
y = -z/3
z = z

seems to satisfy it

Answer 11

http://www.wolframalpha.com/input/?i=%7B%7B1%2C2%2C1%7D%2C%7B3%2C0%2C1%7D%2C%7B1%2C-1%2C0%7D%7D+dot+%7B-n%2F3%2C-n%2F3%2Cn%7D

Answer 12

I think that makes sense, thanks a lot.

Answer 13

youre welcome, and good luck .... linear algebra was kind of a pain when i took it :)

Answer 14

Haha thanks.