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chrisplusian

  • one year ago

Linear algebra question.... see attachment please

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  1. chrisplusian
    • one year ago
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  2. IrishBoy123
    • one year ago
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    .

  3. chrisplusian
    • one year ago
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    Can you see the attachment?

  4. thomas5267
    • one year ago
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    Yes it is possible.

  5. anonymous
    • one year ago
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    Let's call the matrix on the RHS \(M\), so you have \[{\bf{AB}}={\bf{M}}\] If \(|{\bf{A}}|\neq0\), you can find its inverse \({\bf{A}}^{-1}\), which is pretty straightforward since it's 2x2. Then \[{\bf{B}}={\bf{A}}^{-1}{\bf{M}}\] Then you can find the inverse of \(\bf{B}\), provided it's not singular.

  6. chrisplusian
    • one year ago
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  7. thomas5267
    • one year ago
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    Both \(\bf{AB}\) and \(\bf{A}\) have full rank.

  8. chrisplusian
    • one year ago
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    ^^^ that was my attempt but it did not work out

  9. thomas5267
    • one year ago
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    \[ \det(\mathbf{A})=5-(-3)(-3)=5-9=-4\neq5+9=13 \]

  10. chrisplusian
    • one year ago
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    I see that now!!! is my approach correct?

  11. thomas5267
    • one year ago
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    Yes I think.

  12. chrisplusian
    • one year ago
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    Ok thank you

  13. thomas5267
    • one year ago
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    \[ \mathbf{A}^{-1}\text{ and }\mathbf{AB}^{-1}\text{ exist.}\\ \mathbf{A}^{-1}\mathbf{AB}=\mathbf{B}\\ \mathbf{B}^{-1}=\left(\mathbf{A}^{-1}\mathbf{AB}\right)^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A}\\ \left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)\mathbf{B}=\left(\mathbf{AB}\right)^{-1}\left(\mathbf{A}\mathbf{B}\right)=\mathbf{I}\\ \mathbf{B}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\mathbf{AB}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\left(\mathbf{AB}\left(\mathbf{AB}\right)^{-1}\right)\mathbf{A}=\mathbf{I}\\ \therefore \mathbf{B}^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A} \]

  14. chrisplusian
    • one year ago
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    ^^^ I wish I could see it that way.

  15. chrisplusian
    • one year ago
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    If I could give you another medal for that answer I would lol

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