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DLS Group Title

Matrix question!

  • one year ago
  • one year ago

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  1. DLS Group Title
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    What is the adjoint of an adjoint matrix?

    • one year ago
  2. DLS Group Title
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    Is it the original matrix?

    • one year ago
  3. shahara12 Group Title
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    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
  4. cap_n_crunch Group Title
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    It depends on your matrix for a \(2 \times 2\) matrix, it's the original matrix.

    • one year ago
  5. DLS Group Title
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    How and why?What about 3x3 one?why not?

    • one year ago
  6. DLS Group Title
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    @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
  7. cap_n_crunch Group Title
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    For any other \(n \times n\) matrix it's \(\det(A)^{n-2} A\)

    • one year ago
  8. DLS Group Title
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    Its 3x3

    • one year ago
  9. DLS Group Title
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    I am given the adjoint of a matrix I want to find the original matrix.

    • one year ago
  10. cap_n_crunch Group Title
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    So if it's \(3 \times 3\) i could get slightly nasty :)

    • one year ago
  11. cap_n_crunch Group Title
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    But we know that \(\det(\text{adj}(A)) = \det(A)^{n-1}\)

    • one year ago
  12. cap_n_crunch Group Title
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    Which, for your case would mean: \[ \det(A) = \sqrt{\det(\text{adj}(A))} \]

    • one year ago
  13. DLS Group Title
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    :/

    • one year ago
  14. cap_n_crunch Group Title
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    Okay lets call your matrix \[\text{adj}(A) = G\]

    • one year ago
  15. cap_n_crunch Group Title
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    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: \[ A = \frac{1}{\sqrt{\det(G)}} \text{adj}(G) = \frac{1}{\sqrt{\det(G)}} \text{adj}(\text{adj}(A)) \]

    • one year ago
  16. DLS Group Title
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    oh..well.....

    • one year ago
  17. cap_n_crunch Group Title
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    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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