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DLS
Matrix question!
What is the adjoint of an adjoint matrix?
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
It depends on your matrix for a \(2 \times 2\) matrix, it's the original matrix.
How and why?What about 3x3 one?why not?
@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/
For any other \(n \times n\) matrix it's \(\det(A)^{n-2} A\)
I am given the adjoint of a matrix I want to find the original matrix.
So if it's \(3 \times 3\) i could get slightly nasty :)
But we know that \(\det(\text{adj}(A)) = \det(A)^{n-1}\)
Which, for your case would mean: \[ \det(A) = \sqrt{\det(\text{adj}(A))} \]
Okay lets call your matrix \[\text{adj}(A) = G\]
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: \[ A = \frac{1}{\sqrt{\det(G)}} \text{adj}(G) = \frac{1}{\sqrt{\det(G)}} \text{adj}(\text{adj}(A)) \]
It looks complicated but try to understand it step by step. And if you struggle with anythink let me know.