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DLS

  • 2 years ago

Matrix question!

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  1. DLS
    • 2 years ago
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    What is the adjoint of an adjoint matrix?

  2. DLS
    • 2 years ago
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    Is it the original matrix?

  3. shahara12
    • 2 years 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
    • 2 years ago
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    It depends on your matrix for a \(2 \times 2\) matrix, it's the original matrix.

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

  6. DLS
    • 2 years 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
    • 2 years ago
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    For any other \(n \times n\) matrix it's \(\det(A)^{n-2} A\)

  8. DLS
    • 2 years ago
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    Its 3x3

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

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

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

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

  13. DLS
    • 2 years ago
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    :/

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

  15. cap_n_crunch
    • 2 years 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
    • 2 years ago
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    oh..well.....

  17. cap_n_crunch
    • 2 years 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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