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chrisplusian

  • one year ago

Given that a square matrix A satisfies the equation A^2 -7A -I=0 Show that A is invertible and find its inverse.

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  1. Empty
    • one year ago
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    is A a 2x2 matrix?

  2. chrisplusian
    • one year ago
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    It doesn't say that

  3. chrisplusian
    • one year ago
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    I typed the problem verbatim

  4. chrisplusian
    • one year ago
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    \[A^2-7A-I=0\]

  5. Empty
    • one year ago
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    If it's not, I don't think this is enough to imply that it's invertible, since every matrix satisfies its own characteristic equation this is essentially the characteristic equation of a 2x2 matrix, which means it has two distinct eigenvalues. But I think if it's say a 3x3 this isn't really enough.

  6. Empty
    • one year ago
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    Maybe I'm over thinking this though and there's like some easier way to go about this haha

  7. chrisplusian
    • one year ago
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    I don't know what eigenvalues are, and I have asked my professor about the eigenvalues before when someone on here helped me and started by trying to explain them. He said that until he explains it to us the solutions to our problems won't require comprehending eigenvalues.

  8. Empty
    • one year ago
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    Ok, I think I found something better then for you, we don't need to go so deep, sorry about that!

  9. chrisplusian
    • one year ago
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    What we have covered are matrix algebra, matrix inverses, vectors, transpose of a matrix, lines and planes in this linear algebra course

  10. Empty
    • one year ago
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    Ok what we have to do is factor it and rearrange this equation, that will give you the inverse: \[A^2-7A-I=0\] here I add the identity to both sides and rewrite \(A=AI\) which is fine since a matrix times the identity is just itself. \[A^2-7AI=I\] Now I can factor out one of the A's \[A(A-7I)=I\] Oh hey, a matrix times another matrix equal to the identity matrix means that it must be its inverse! \[AA^{-1}=I\] So here you go: \[A^{-1}=A-7I\]

  11. chrisplusian
    • one year ago
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    Could it be that simple? I must be overthinking it too. I was thinking they wanted specific matrix components...... I was wondering how I was even supposed to decide what size matrix it was. Thanks

  12. chrisplusian
    • one year ago
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    ONe thing....

  13. Empty
    • one year ago
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    Haha yeah, if there's any step in particular that you seem doubtful about I could help explain more, yeah sure anything.

  14. chrisplusian
    • one year ago
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    How does this show that it is invertible?

  15. chrisplusian
    • one year ago
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    Because at one point it is equal to the identity matrix?

  16. Empty
    • one year ago
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    Ok, so suppose I show you this matrix equation, \[AB=I\] Is B the inverse of A?

  17. chrisplusian
    • one year ago
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    Or would I need to show that multiplication both ways results in the identity matrix?

  18. chrisplusian
    • one year ago
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    AB=I, and BA=I is how you prove it is invertible right?

  19. Empty
    • one year ago
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    Yeah you could show that if you like, but once you know \[AB=I\] then you know \[BA=I\] since without even knowing anything we can prove it. In this example if you want to show that, you can do that if you like as well, just do what I've done but factor out A on the right instead. I'll give the proof that \[AB=I\] implies that \[BA=I\] in my next post.

  20. chrisplusian
    • one year ago
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    I actually looked back at the work I did before I posted this question and I have \[(A-7)A=I\] does that seem like it is ok without the identity attached to the 7?

  21. Empty
    • one year ago
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    Not quite, because what does it mean to add a scalar to a matrix?

  22. Empty
    • one year ago
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    However it is common to just represent a scalar multiplied by an identity matrix as a so-called 'scalar matrix' so as long as you understand that the 7 you have written there represents a matrix with 7's down the diagonal you're fine.

  23. chrisplusian
    • one year ago
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    I know this sounds silly but what does it mean to add a scalar to a matrix? You can't do that right?

  24. Empty
    • one year ago
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    Yeah you can't! That's why you have to have the identity matrix there, so really that 7 you've written is shorthand: \[7 \implies 7I\]

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