DLS
  • DLS
Matrix question!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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DLS
  • DLS
What is the adjoint of an adjoint matrix?
DLS
  • DLS
Is it the original matrix?
anonymous
  • anonymous
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

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

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