Please help, do not know what this type of problem is called or how to solve it!

  • #1
Suppose that A is an nxn matrix which satisfies the equation

A3 - 2A2 + 3A - I = 0

Show that the nxn matrix B defined by the equation B = A2 - 2A +3I is invertible and find its inverse.

Does anyone have any idea what this type of problem is called, and what steps you take to solve it?
 

Answers and Replies

  • #2
hotvette
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It's a matrix algebra (linear algebra) problem. Try factoring it like you would a normal algebra problem. You'll also need to recall the definition of a matrix inverse to solve the problem.
 
  • #3
can you be more specific? for some reason i just can't see how to do this, i've literally spent over 2 hours attempting to solve it.
 
  • #4
hotvette
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You need to tell us what you've tried and where you are stuck.
 
  • #5
i think what's messing me up is the fact that there are equations involved. i understand how to do the inverse of a matrix, but I can't seem to connect the dots between the idea of a matrix and how it relates to the given equations.
 
  • #6
jbunniii
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i think what's messing me up is the fact that there are equations involved. i understand how to do the inverse of a matrix, but I can't seem to connect the dots between the idea of a matrix and how it relates to the given equations.

Try to recognize how the two equations are related to each other.
 
  • #7
hotvette
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OK, so the conceptual block is the fact that matrices can be combined with operators to make expressions. The rules are almost the same as with normal algebra. The following is a matrix algebra equation that says a matrix B added to a matrix A equals matrix C:

A + B = C

Matrices can also be multiplied by scalars (meaning each element of the matrix is multiplied by the scalar):

A + 2B = C

Here are examples of an expressions using matrix multiplication (and addition):

AB = C
AB + C = D
A2 = AA
AB = I

In the last case, what can you say about B?

Also, matrix mutliplication is distributive:

A(B + C) = AB + AC

Is none of the above familiar?
 
  • #8
I recognize all that. In the last case, B would be equal to the inverse of A. I know there is just something so small that I am missing, and I know I'll feel like an idiot when I see it. do i have to solve for A in the first equation so that I have an A= and a B=? idk i'm just frustrated, this stuff usually comes so easily to me, for the first time I understand what it's like to not be good at math.
 
  • #9
hotvette
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Suggestion: solve the first equation for I and factor the opposite side.
 
  • #10
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Let's just abuse some notation for a minute and say [itex]I[/itex] is [itex]1[/itex]. Carrying this out would give you

[tex]\begin{align*}
& A^3 - 2A^2 +3A - 1 = 0 \\
& A^2 - 2A + 3 = B
\end{align*}
[/tex]

Do you think you can make a simple substitution here?
 
  • #11
you can factor out an A from the first equation and be left with A(B) = 1, am I on the right track?
 
  • #12
hotvette
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Yes. And what does that tell you about B?
 
  • #13
that it's equal to the inverse of A.
 
  • #14
hotvette
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Yep. Now you should be in a position to answer the original questions.
 
  • #15
okay I wanna say I get it but I think my issue here is I don't know what format answer I am looking for. Is the inverse just going to be in the form of an equation as well?
 
  • #16
Ray Vickson
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okay I wanna say I get it but I think my issue here is I don't know what format answer I am looking for. Is the inverse just going to be in the form of an equation as well?

What is wrong with saying that the inverse of B is A? After all, that is what you have just shown, and it does answer the original question completely.

RGV
 
  • #17
yeah that would work, I guess it's one of those times the simplest answer is the right answer. thanks for all your help.
 
  • #18
HallsofIvy
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From [itex]A^3 - 2A^2 + 3A - I = 0[/itex] we can get [itex]A^3- 2A^2+ 3A= I[/itex] and then [itex]A(A^2- 2A+ 3)= I[/itex]. Do you see the point?
 
  • #19
Curious3141
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you can factor out an A from the first equation and be left with A(B) = 1, am I on the right track?

Yes, you're on the right track, although that should be AB = I.

Now postmultiply both sides by [itex]B^{-1}[/itex].
 

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