# Determinant and trace of matrix ( HELP)

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

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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.

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

What do you know about eigenvalues? Do you know what the characteristic polynomial is?

Not much at the moment im afraid? I dont know what the characteristic polynomial is??

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...

Ok so by I u mean an identity matrix??

Yes, I is the identity matrix!

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?

Yes, substract those two matrices, and then solve the associated system of equations...

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??

Yes, this is correct. So, what does a determinant 0 tell you?

Erm that there exists a non zero solution?

Yes, so you have shown that 6 is an eigenvalue!

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???

Yes, that is correct. In fact, the polynomial $$\lambda^2-4\lambda+3$$ is called the characteristic polynomial. It seems that you came up with that concept by yourself! Haha ok thanks micromass.

I have one final question micromass, using the fact 6 is an eigenvalue and the determinant how would I find the remaining eigenvalues?