Determinant and trace of matrix ( HELP)

  • Thread starter andrey21
  • Start date
  • #1
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FInd the determinant of the following matrix?

4,-4,-8
-2, 2, 6
0, 0,-1




Heres my attempt

4.(2x(-1)- 6x0) -(-4).((-2)x(-1) - 6x0) +(-8).((-2)x0-2x0)

which goves:

4.(-2)+4(2) -8 = 0

is this correct??

Im also asked to find the trace? What is this and how do i find it?
Thanks
 

Answers and Replies

  • #2
22,089
3,293
Yes, this is correct. However, it could have been done easier if you had take the third row to calculate the determinant. Then you would see immediately that the determinant is (-1)(4.2-(-2)(-4))=0.

The trace is simply the sum of the diagonal elements.
 
  • #3
466
0
Ah ok Thanks micromass. For the trace I obtained:

4+2-1 = 5 Which is simple enough.

Now Im asked to show that 6 is an eigenvalue of the matrix. How would I go about doing that? This is a new topic for me so im struggling a little. Thnks
 
  • #4
22,089
3,293
What do you know about eigenvalues? Do you know what the characteristic polynomial is?
 
  • #5
466
0
Not much at the moment im afraid? I dont know what the characteristic polynomial is??
 
  • #6
22,089
3,293
Then you'll just have to do it the hard way. Let A be our matrix. You'll need to show that there exists a vector x such that Ax=6x. This is equivalent to saying that (A-6I)x=0. Thus you must show that the system (A-6I)x=0 has a non-zero solution...
 
  • #7
466
0
Ok so by I u mean an identity matrix??
 
  • #8
22,089
3,293
Yes, I is the identity matrix!
 
  • #9
466
0
So I have to basically subtract an identity matrix of:

6,0,0
0,6,0
0,0,6

as it is 6I from my matrix??

Im I on the rite track?
 
  • #10
22,089
3,293
Yes, substract those two matrices, and then solve the associated system of equations...
 
  • #11
466
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ok from that I get the matrix:

-2,-4,-8
-2,-4, 6
0, 0,-7

using previous determinant method I obtain:

-2(28) + 4(14) -8(0)

= 0

is this correct??
 
  • #12
22,089
3,293
Yes, this is correct. So, what does a determinant 0 tell you?
 
  • #13
466
0
Erm that there exists a non zero solution?
 
  • #14
22,089
3,293
Yes, so you have shown that 6 is an eigenvalue!
 
  • #15
466
0
Brilliant thanks micromass. Just another question say Im asked to find the eigen value of the following matrix:

2,1
1,2

Do I simply do (A-lambda I) = det (2-Lambda 1
1 1-Lambba)

giving me:

(lambda)^2 -4(Lambda) +3 = 0

lambda = 1 lambda = 3

which are the eigenvalues???
 
  • #16
22,089
3,293
Yes, that is correct. In fact, the polynomial [tex]\lambda^2-4\lambda+3[/tex] is called the characteristic polynomial. It seems that you came up with that concept by yourself! :smile:
 
  • #17
466
0
Haha ok thanks micromass.
 
  • #18
466
0
I have one final question micromass, using the fact 6 is an eigenvalue and the determinant how would I find the remaining eigenvalues?
 

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