What is the Notation for Adjoints in Matrices?

[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:

det(A^-1 + adj(A))

Thanks in advance.

 

[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.

 

[courtrigrad]

isn't adj(A) the adjoint matrix?

 

[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)

 

[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
A*Adj(A) = det(A)*I
det(Adj(A)) = det(A)^(n-1)
det(adj(A)) = 2^4 = 16

 

[Don Aman]

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

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

 

[cronxeh]

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

of course!

 

[neo143]

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

 

[ac_nex]

Tough question!

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

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

Any usefull remarks.

 

[neo143]

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

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

 

[Muzza]

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

 

[cronxeh]

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

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

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

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

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

Edit: you already got it

 

[ac_nex]

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

 

[mathwonk]

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

i.e. since adj(A).det(A) = A^(-1),

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

hence det(A^(-1)+adj(A)) = 3^5/2.

 

[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?