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DLS
 one year ago
Matrix question!
DLS
 one year ago
Matrix question!

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

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

shahara12
 one year 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

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

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

DLS
 one year 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/

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

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

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

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

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

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

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

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