Tricky one - solve for k - linear algebra

Hi sciencegirl1!

doesn't look right …

are you sure you've copied the question correctly?

 

If x is an eigenvector, Ax=qx, for some constant q, which when multiplied out, will give you three equations in two variables. Unfortunately, when I tried to work it out, I got 23=3q and 13=5q, which leads me to believe that [3, 5, 2] isn't an eigenvector of this matrix.

 

D'uh!

ok … now just do the obvious …

multiply the vector by the matrix … what do you get?

 

Hi sciencegirl1!

I can't help thinking that you try to make everything too difficult …

go back to basics …

Q: what is the definition of an eigenvector? A: if you multiply it by the matrix, the result is an exact multiple of the original

sooooo … what value of k makes -15 -11+3k 9 an exact multiple of -5 -2 3 ?

 

oh, its 5k now, is it? :rofl:

well, that does make it a lot easier

nooo, it's not -2 (and how do you get 5/3 out of that anyway? )

Hint: you'll make less mistakes if you do it in stages …

so: stage 1: what is the eigenvalue?

 

Nope!

You try it … what is the eigenvalue?

 

Some are using the vector and/or matrix you gave originally, not the corrected ones which are now buried in the middle of many posts.

You are given

[tex]
A=
\begin{pmatrix}
2 & 1 & -1 \\
1 & 3 & k \\
-1 & -5 & -2
\end{pmatrix}
\ \ \ \ \ \
x=
\begin{pmatrix}
-5 \\
-2 \\
3
\end{pmatrix}
[/tex]

and are told that [itex]x[/itex] is an eigenvalue of [itex]A[/itex], which means that

[tex]Ax = \lambda x[/tex]

where on the left, we matrix-multiply the matrix [itex]A[/itex] times the column-vector [itex]x[/itex] to get a new column vector; some of the components of this column vector will depend on [itex]k[/itex]. On the right, we multiply each component of [itex]x[/itex] by the number [itex]\lambda[/itex] (which we don't know yet). We equate each component of the left side with the corresponding component of the right side. This gives us three equations in two unknowns ([itex]k[/itex] and [itex]\lambda[/itex]). Presumably, the equations will be compatible; that is, there will be a solution. So, pick any two equations, solve them for [itex]k[/itex] and [itex]\lambda[/itex], and then, if you like, check that the third equation is also satisfied.

 

You are trying to make the individual vector components match up, the sum of the components is fairly meaningless here.

 

No, each component of each side is equal.

First of all,

[tex]\lambda\times
\begin{pmatrix}
-5 \\
-2 \\
3
\end{pmatrix}
= \begin{pmatrix}
-5\lambda \\
-2\lambda \\
3\lambda
\end{pmatrix}
[/tex]

Then, the left side equals the right side:

[tex]
\begin{pmatrix}
-15 \\
-11+3k \\
9
\end{pmatrix} =
\begin{pmatrix}
-5\lambda \\
-2\lambda \\
3\lambda
\end{pmatrix}[/tex]

This gives us three equations, one for each component:

[tex]-15 = -5\lambda[/tex]

[tex]-11+3k = -2\lambda[/tex]

[tex]9 = 3\lambda[/tex]