1. Nov 20, 2005

### ac_nex

Just wondering, how would you solve a problem such as this one:

Suppose A is an 5 x 5 matrix, with det(A) = 2 find the following:

2. Nov 20, 2005

### mathwonk

my first instinct is to multiply by A to see what happens, but I don't know the definition of adj(A). so I don't know what adj(A).A is.

then use the fact that det respects products.

3. Nov 20, 2005

4. Nov 20, 2005

### Don Aman

if adj means the adjugate, that is the transpose of the matrix of minors (which I guess some people also call the adjoint, but I save that for the Hermitian adjoint), then use the fact that A^-1=adj(A)/det(A)

5. Nov 20, 2005

### cronxeh

A*A^-1 = I

det(A*A^-1)=det(I)=5
det(A)*det(A^-1)=5
2*det(A^-1)=5
det(A^-1)=5/2

6. Nov 21, 2005

### Don Aman

I'm pretty sure that det(1)=1

7. Nov 21, 2005

### cronxeh

of course!

8. Nov 21, 2005

### neo143

det(A)=2 given
A^-1 + adj(A) = A^-1 + 2 A^-1
=3 A^-1
det (3A^-1) = 3 det(A^-1) = 3 (det(A))^-1
=3*2^-1
=3/2

How's This

9. Nov 21, 2005

### ac_nex

Tough question!

Actually guys, thanks for considering my question, but Im afraid all of your answers are different from what the actual answer is.

It says the answer is (3^5)/2

Any usefull remarks.

10. Nov 21, 2005

### neo143

Sorry in my earlier solution there was a problem
This is the correct solution

det(A)=2 given
A^-1 + adj(A) = A^-1 + 2 A^-1
=3 A^-1
=>det (3A^-1)
since A is 5*5 matrix
det(3A^-1)=3^5 det(A^-1)
=3^5 (det A)^-1
=3^5 (2)^-1
=3^5/2

enjoy

11. Nov 21, 2005

### Muzza

neo143 says that det (3A^-1) = 3 det(A^-1), which is incorrect. You should be able to fix this easily yourself, though.

12. Nov 21, 2005

### cronxeh

det(a*A) = (a^n)*det(A)

A^-1 = adj(A)/det(A) , adj(A)=(1/2)*(A^-1) , A*A^-1 = I , det(A*A^-1)=1=det(A)*det(A^-1) , det(A^-1) = 1/2

det(3/2*adj(A)) = ( (3/2)^5 ) * (( 2^5))/2 = 121.5

13. Nov 21, 2005

### ac_nex

Thanks, I already knew how to do it after some serious thinking about neo143's first post. Thanks again.

14. Nov 22, 2005

### mathwonk

i think i completely solved it for you, modulo the definition of adj.

my advice gives det(A)det(A^(-1)+adj(A)) = det(I + 2I) = det(3I) = 3^5.

15. Nov 24, 2005

### QMrocks

how many definitions does adjoint take?

1) there is the classical adjoint (its exact definition too messy to write) which has the useful relation A^(-1)=Adj(A)/det(A).

2) then there is the definition of adjoint as the transpose and conjugate of a matrix.

These two adjoint operation are different. May i know what notation is usually used? Adj?