# Determinant and trace of matrix ( HELP)

andrey21
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

Staff Emeritus
Homework Helper
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.

andrey21
Ah ok Thanks micromass. For the trace I obtained:

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

Now I am 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 I am struggling a little. Thnks

Staff Emeritus
Homework Helper
What do you know about eigenvalues? Do you know what the characteristic polynomial is?

andrey21
Not much at the moment I am afraid? I don't know what the characteristic polynomial is??

Staff Emeritus
Homework Helper
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...

andrey21
Ok so by I u mean an identity matrix??

Staff Emeritus
Homework Helper
Yes, I is the identity matrix!

andrey21
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?

Staff Emeritus
Homework Helper
Yes, substract those two matrices, and then solve the associated system of equations...

andrey21
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??

Staff Emeritus
Homework Helper
Yes, this is correct. So, what does a determinant 0 tell you?

andrey21
Erm that there exists a non zero solution?

Staff Emeritus
Homework Helper
Yes, so you have shown that 6 is an eigenvalue!

andrey21
Brilliant thanks micromass. Just another question say I am 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?

Staff Emeritus
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!