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DLS

  • one year ago

Matrix question!

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  1. DLS
    • one year ago
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    What is the adjoint of an adjoint matrix?

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

  3. shahara12
    • one year ago
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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

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

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

  6. DLS
    • one year ago
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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/

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

  8. DLS
    • one year ago
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    Its 3x3

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

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

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

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

  13. DLS
    • one year ago
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    :/

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

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

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

  17. cap_n_crunch
    • one year ago
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    It looks complicated but try to understand it step by step. And if you struggle with anythink let me know.

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