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DLS
 3 years ago
Matrix question!
DLS
 3 years ago
Matrix question!

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DLS
 3 years ago
Best ResponseYou've already chosen the best response.1What is the adjoint of an adjoint matrix?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It depends on your matrix for a \(2 \times 2\) matrix, it's the original matrix.

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

DLS
 3 years ago
Best 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/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For any other \(n \times n\) matrix it's \(\det(A)^{n2} A\)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1I am given the adjoint of a matrix I want to find the original matrix.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So if it's \(3 \times 3\) i could get slightly nasty :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But we know that \(\det(\text{adj}(A)) = \det(A)^{n1}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Which, for your case would mean: \[ \det(A) = \sqrt{\det(\text{adj}(A))} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay lets call your matrix \[\text{adj}(A) = G\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We 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: \[ A = \frac{1}{\sqrt{\det(G)}} \text{adj}(G) = \frac{1}{\sqrt{\det(G)}} \text{adj}(\text{adj}(A)) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It looks complicated but try to understand it step by step. And if you struggle with anythink let me know.
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