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DLS Group TitleBest ResponseYou've already chosen the best response.1
What is the adjoint of an adjoint matrix?
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
Is it the original matrix?
 one year ago

shahara12 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
It depends on your matrix for a \(2 \times 2\) matrix, it's the original matrix.
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
How and why?What about 3x3 one?why not?
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
@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/
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
For any other \(n \times n\) matrix it's \(\det(A)^{n2} A\)
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
I am given the adjoint of a matrix I want to find the original matrix.
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
So if it's \(3 \times 3\) i could get slightly nasty :)
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
But we know that \(\det(\text{adj}(A)) = \det(A)^{n1}\)
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
Which, for your case would mean: \[ \det(A) = \sqrt{\det(\text{adj}(A))} \]
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
Okay lets call your matrix \[\text{adj}(A) = G\]
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
We know: \[ \underbrace{\text{adj}(G)}_{\text{adj}(\text{adj}(A))} = \det(A)^{n2}A \] and \[ \det(G) = \det(A)^{n1} \] So in your case it's \(3 \times 3\). Therefore we can write: \[ \det(G) = \det(A)^{31} = \det(A)^2 \] Which makes\[ \det(A) = \sqrt{\det(G)} \] Now we're coming back to our first formula: \[ \text{adj}(G) = \det(A)^{n2} A \] We plug in the formula we got for \(\det(A):\)\[ \text{adj}(G) = \sqrt{\det(G)}^{n2} A \] 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: \[ A = \frac{1}{\sqrt{\det(G)}} \text{adj}(G) = \frac{1}{\sqrt{\det(G)}} \text{adj}(\text{adj}(A)) \]
 one year ago

cap_n_crunch Group TitleBest ResponseYou've already chosen the best response.1
It looks complicated but try to understand it step by step. And if you struggle with anythink let me know.
 one year ago
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