This discussion is now closed.

Check out other Related discussions

- System of equations question
- Vectors Help
- A level chemistry
- Chemistry IA.
- BTEC applied science unit 19
- gcse romeo and juliet essay, could someone mark it please!
- Limestone
- A Level Physics
- Question on A level grade boundaries?
- Grades
- AQA A Level chemistry
- could anybody mark my romeo and juliet essay?
- Chemistry Calculation Question
- OCR B (MEI) AS Further Maths Exam Q8 (2020) - MATRIX DETERMINANT AND REFLECTION
- Chemistry Oxford 2nd interview
- Geometry in curved space
- Classification
- Eduqas A level RE paper 3 ethics 2023
- AQA A level Psychology
- University applications

how do i find det(adj A) ?

where A is an nxn matrix.

thanks xxxxxxxx

where A is an nxn matrix.

thanks xxxxxxxx

dvs

You know that A adj(A) = det(A) I, so take determinants:

det(A) det(adj(A)) = (det(A))^n

(I used det(AB)=det(A)det(B) and det(cA)=c^ndet(A).)

So if A is invertible then det(adj(A)) = (det(A))^(n-1). Otherwise det(adj(A))=0. Hence in both cases det(adj(A)) = (det(A))^(n-1).

det(A) det(adj(A)) = (det(A))^n

(I used det(AB)=det(A)det(B) and det(cA)=c^ndet(A).)

So if A is invertible then det(adj(A)) = (det(A))^(n-1). Otherwise det(adj(A))=0. Hence in both cases det(adj(A)) = (det(A))^(n-1).

thanks that is helpful. one thing, where did you get the c^n from?

dvs

If you multiply each of the n rows by c, then you multiply the determinant by c n times.

ok. but where does the c come from?!

A^{-1 }= 1/det(A) * adj(A)

Take determinant of both sides

det(A^{-1}) = det(1/det(A) * adj(A))

Simplify

det(A^{-1}) = 1/det(A) * det(adj(A))

det(A) * det(A^{-1}) = det(adj(A))

So as you can see, you just need to find det(A) in order to know det(adj(A))

Take determinant of both sides

det(A

Simplify

det(A

det(A) * det(A

So as you can see, you just need to find det(A) in order to know det(adj(A))

(edited 9 years ago)

Original post by TeslerCoil

A^{-1 }= 1/det(A) * adj(A)

Take determinant of both sides

det(A^{-1}) = det(1/det(A) * adj(A))

Simplify

det(A^{-1}) = 1/det(A) * det(adj(A))

det(A) * det(A^{-1}) = det(adj(A))

So as you can see, you just need to find det(A) in order to know det(adj(A))

Take determinant of both sides

det(A

Simplify

det(A

det(A) * det(A

So as you can see, you just need to find det(A) in order to know det(adj(A))

Original post by DFranklin

Note that the last post in this thread before yours is 8 years old. I suspect they are no longer waiting for an answer...

As OP reads it, a single tear rolls down their cheek. "If this answer had come just a week earlier, the world could have been saved!" they whisper, as across the globe the fires begin.

Original post by Smaug123

As OP reads it, a single tear rolls down their cheek. "If this answer had come just a week earlier, the world could have been saved!" they whisper, as across the globe the fires begin.

- System of equations question
- Vectors Help
- A level chemistry
- Chemistry IA.
- BTEC applied science unit 19
- gcse romeo and juliet essay, could someone mark it please!
- Limestone
- A Level Physics
- Question on A level grade boundaries?
- Grades
- AQA A Level chemistry
- could anybody mark my romeo and juliet essay?
- Chemistry Calculation Question
- OCR B (MEI) AS Further Maths Exam Q8 (2020) - MATRIX DETERMINANT AND REFLECTION
- Chemistry Oxford 2nd interview
- Geometry in curved space
- Classification
- Eduqas A level RE paper 3 ethics 2023
- AQA A level Psychology
- University applications

Latest

Trending

Last reply 4 days ago

can someone please explain what principle domain is and why the answer is a not c?Maths

0

13