Question

Matrix question!

Answer 1

What is the adjoint of an adjoint matrix?

Answer 2

Is it the original matrix?

Answer 3

The matrix formed by taking the transpose of the cofactor matrix of a given original matrix. The adjoint of matrix A is often written adj A

Answer 4

It depends on your matrix for a \(2 \times 2\) matrix, it's the original matrix.

Answer 5

How and why?What about 3x3 one?why not?

Answer 6

@shubhamsrg That solves the IIT Main question,we had to take the adjoint of that ajdoint matrix then finally the determinant then equate it to the given value If i'm not wrong/

Answer 7

For any other \(n \times n\) matrix it's \(\det(A)^{n-2} A\)

Answer 8

Its 3x3

Answer 9

I am given the adjoint of a matrix I want to find the original matrix.

Answer 10

So if it's \(3 \times 3\) i could get slightly nasty :)

Answer 11

But we know that \(\det(\text{adj}(A)) = \det(A)^{n-1}\)

Answer 12

Which, for your case would mean: \[
\det(A) = \sqrt{\det(\text{adj}(A))}
\]

Answer 13

:/

Answer 14

Okay lets call your matrix \[\text{adj}(A) = G\]

Answer 15

We know: \[
\underbrace{\text{adj}(G)}_{\text{adj}(\text{adj}(A))}
= \det(A)^{n-2}A
\]
and
\[
\det(G) = \det(A)^{n-1}
\]
So in your case it's \(3 \times 3\). Therefore we can write:
\[
\det(G) = \det(A)^{3-1} = \det(A)^2
\]
Which makes\[
\det(A) = \sqrt{\det(G)}
\]
Now we're coming back to our first formula:
\[
\text{adj}(G) = \det(A)^{n-2} A
\]
We plug in the formula we got for \(\det(A):\)\[
\text{adj}(G) = \sqrt{\det(G)}^{n-2} A
\]
With \(n = 3\) it simplifies to:
\[
\text{adj}(G) =\sqrt{\det(G)} A
\]
Finally, to get \(A\) we multiply \(\frac{1}{\sqrt{\det(G)}}\) on both sides:
\[
\text{adj}(G) \frac{1}{\sqrt{\det(G)}} =\sqrt{\det(G)}\frac{1}{\sqrt{\det(G)}} A
\]
We can do some cancellations and end up with:
\[
A = \frac{1}{\sqrt{\det(G)}} \text{adj}(G) = \frac{1}{\sqrt{\det(G)}} \text{adj}(\text{adj}(A))
\]

Answer 16

oh..well.....

Answer 17

It looks complicated but try to understand it step by step. And if you struggle with anythink let me know.