# Thread: linear algebra- upside down y symbol??

1. ## linear algebra- upside down y symbol??

For the matrix
A =

[6 3 3
3 6 3
3 3 6]

determine all possible values of ** such that det(A − **I3) = 0.

NOTE...where the ** are there is a symbol like an upside down y.
what is the question asking for??

2. Originally Posted by linearalgebra
For the matrix
A =

[6 3 3
3 6 3
3 3 6]

determine all possible values of ** such that det(A − **I3) = 0.

NOTE...where the ** are there is a symbol like an upside down y.
what is the question asking for??
I can't help you with the real question, but I think with an upside-down y you mean a lambda. written as λ and used in Eigenvalue, eigenvector and eigenspace - Wikipedia, the free encyclopedia .

3. i just found that out.
unfortunately i still dont get how to find it out.
i'll try to find some explanation on wikipedia or something..thanks though

4. Originally Posted by linearalgebra
For the matrix
A =

[6 3 3
3 6 3
3 3 6]

determine all possible values of ** such that det(A − **I3) = 0.

NOTE...where the ** are there is a symbol like an upside down y.
what is the question asking for??
I assume $\lambda$ is the symbol to which you refer. It is the greek letter "lambda". Now onto your question, in matrices $\lambda$ is used to represent the eigenvalues of a matrix. If you've not come across this term then I assume you will soon as the question is basically asking you to find these eigenvalues so for now just think of them as numbers (that's all they are really).
First step is to construct the matrix $A-\lambda I$ so you will have something like $\left[\begin{array}{ccc}6-\lambda & 3 & 3\\ 3 & 6-\lambda & 3\\ 3 & 3 & 6-\lambda\end{array}\right]$
Now all you have to do is find its determinant which will be a cubic polynomial and then as the question suggests put it equal to zero and find its roots.
Is this enough to get you started?
Hopefully it was just the $\lambda$ that was throwing you off.

5. thanks...clears up a little but im still unsure as to how to go about solving this.

when you say..."find its determinant which will be a cubic polynomial" what do you mean?

sorry if i sound totally oblivious, ive been looking at this for so long that i think ive confused myself

6. Originally Posted by linearalgebra
thanks...clears up a little but im still unsure as to how to go about solving this.

when you say..."find its determinant which will be a cubic polynomial" what do you mean?

sorry if i sound totally oblivious, ive been looking at this for so long that i think ive confused myself
No problem, what I mean is that when you take the det you will get an equation which looks like
$a\lambda^{3}+b\lambda^{2}+c\lambda+d=0$ and you will have to factorise (if it factorises) this to find values of lambda.

7. okay, so here is what i gather...

by using the following formula which is the determinant of a three by three matrix is

a11 a12 a13
a21 a22 a23
a31 a32 a33

= a11(a22a33 − a23a32) − a12(a21a33 − a23a31) + a13(a21a32 − a22a31)

three steps
1) I use this formula and apply it to my original matrix
2) I use that formula and apply it to the 3x3 identity matrix
3) the values i get from those, i plus in to the question.

am i correct?

8. Originally Posted by linearalgebra
okay, so here is what i gather...

by using the following formula which is the determinant of a three by three matrix is

a11 a12 a13
a21 a22 a23
a31 a32 a33

= a11(a22a33 − a23a32) − a12(a21a33 − a23a31) + a13(a21a32 − a22a31)

three steps
1) I use this formula and apply it to my original matrix
2) I use that formula and apply it to the 3x3 identity matrix
3) the values i get from those, i plus in to the question.

am i correct?
For a 3x3 matrix of the form

$\left[\begin{array}{ccc}a & b & c\\ d & e & f\\ h & j & k\end{array}\right]$

It's determinant is

$a\left|\begin{array}{cc}e & f\\ j & k\end{array}\right| - b\left|\begin{array}{cc}d & f\\ h & k\end{array}\right| + c\left|\begin{array}{cc}d & e\\ h & j\end{array}\right|$

$= a(ek - fj) - b(dk - fh) + c(dj - eh)$.

Apply this to your particular matrix, set it equal to $0$, and solve for $\lambda$.

9. awesome!! i didnt get your formula in time, but its the same thing as the one i posted anyway..
okay so i did it and ended up with A= 108
and I3= 1
therefore making the unknown value 108

this case is closed!
=) thank you so much for your help!

ive got two other problems listed, and im working my way through those as well, but if you wouldnt mind just taking a quick peek at them.

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