Property of Determinants Answers Check

[muzziMsyed21]

Homework Statement

Let A and P be square matrices of the same size with P invertible, Prove detA=det(P^-1AP)

Homework Equations

Suppose that A and B are square matrices of the same size. Then det(AB)=det(A)det(B)

The Attempt at a Solution

detA=det(P^-1AP)
detA=det(P^-1PA)
detA=det(IA)
detA=1*detA
detA=detA

SECOND QUESTION:

Homework Statement

Let A be an nxn matrix. Prove that if matrix A satisfies 7A²+8A+3I=[0]

Homework Equations

Invertible Matrix Theorem

The Attempt at a Solution

7A²+8A+3I=[0]
7A²+8A = -3I
A(7A+8)=-3I

From this point I don't know if I am headed towards the correct answer or have the right idea THIRD QUESTION:

Homework Statement

Suppose A is an nxn matrix satisfying A^T+A=[0], where n is odd. Prove detA=0.

Homework Equations

detA^T=detA

The Attempt at a Solution

A^T+A=[0]
A^T=-A
detA^T=det(-A)
since detA^T=detA
detA=det(-A)

I think I've got the right idea...

 

[Mark44]

Homework Statement

Let A and P be square matrices of the same size with P invertible, Prove detA=det(P^-1AP)

Homework Equations

Suppose that A and B are square matrices of the same size. Then det(AB)=det(A)det(B)

The Attempt at a Solution

detA=det(P^-1AP)
detA=det(P^-1PA)

You can't do this (above). Matrix multiplication is not generally commutative, so in general, AP ≠ PA

detA=det(IA)
detA=1*detA
detA=detA

SECOND QUESTION:

Homework Statement

Let A be an nxn matrix. Prove that if matrix A satisfies 7A²+8A+3A=[0]

Typo in your equation?

Homework Equations

Invertible Matrix Theorem

The Attempt at a Solution

7A²+8A+3A=[0]

Typo?

7A²+8A = -3I
A(7A+8)=-3I

From this point I don't know if I am headed towards the correct answer or have the right idea THIRD QUESTION:

Homework Statement

Suppose A is an nxn matrix satisfying A^T+A=[0], where n is odd. Prove detA=0.

Homework Equations

detA^T=detA

The Attempt at a Solution

A^T+A=[0]
A^T=-A
detA^T=det(-A)
since detA^T=detA
detA=det(-A)

How does this show that det(A) = 0?
Also, what role does the fact that n is odd play?

I think I've got the right idea...

 

[muzziMsyed21]

sorry for the typos I edited and fixed them

 

[muzziMsyed21]

You can't do this (above). Matrix multiplication is not generally commutative, so in general, AP ≠ PA
Typo in your equation?
Typo?
How does this show that det(A) = 0?
Also, what role does the fact that n is odd play?

that is where i am stumped... is it det(A) = -det(A) then detA=0 ?

 

[Dick]

that is where i am stumped... is it det(A) = -det(A) then detA=0 ?

IF det(A)=(-det(A)) then det(A)=0, sure. det(-A) isn't always equal to -det(A). What is det(kA) for k a constant?

 

[muzziMsyed21]

IF det(A)=(-det(A)) then det(A)=0, sure. det(-A) isn't always equal to -det(A). What is det(kA) for k a constant?

kⁿdet(A)

 

[Dick]

kⁿdet(A)

Well, sure. Now put k=(-1) and figure out what n being odd might have to do with it.

 

[muzziMsyed21]

IF det(A)=(-det(A)) then det(A)=0, sure. det(-A) isn't always equal to -det(A). What is det(kA) for k a constant?

det(A^T)=det(-A)
det(A)=(-1)ⁿdetA

from here i still don't know how det=0

 

[Dick]

det(A^T)=det(-A)
det(A)=(-1)ⁿdetA

from here i still don't know how det=0

It only has to be zero if n is odd. Then (-1)^n=(-1), right? det is a real number. If a real number x=(-x) then x=0. Can you prove that?

 

[muzziMsyed21]

It only has to be zero if n is odd. Then (-1)^n=(-1), right? det is a real number. If a real number x=(-x) then x=0. Can you prove that?

does it have anything to do with additive inverse?

 

[Mark44]

does it have anything to do with additive inverse?

Don't overthink this.

Since n is odd, (-1)ⁿ = -1, so
det(A^T) = det(-A) = (-1)ⁿdet(A) = ?
What can you conclude from the above?

 

[Mark44]

Also, it's better to post one problem per thread.

 

[Mark44]

7A₂+8A+3I=[0]
7A²+8A = -3I
A(7A+8)=-3I

The last line should be A(7A + 8I) = -3I
It doesn't make sense to add a number to a matrix.

Rearranging a bit results in this equation.
So A * (-1/3)(7A + 8I) = I

Does this give you any ideas?