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

[steph_swords]

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?

 

[hotvette]

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.

 

[steph_swords]

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.

 

[hotvette]

You need to tell us what you've tried and where you are stuck.

 

[steph_swords]

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.

 

[jbunniii]

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.

 

[hotvette]

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?

 

[steph_swords]

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.

 

[hotvette]

Suggestion: solve the first equation for I and factor the opposite side.

 

[Muphrid]

Let's just abuse some notation for a minute and say $I$ is $1$. Carrying this out would give you

$$\begin{align*} & A^3 - 2A^2 +3A - 1 = 0 \\ & A^2 - 2A + 3 = B \end{align*}$$

Do you think you can make a simple substitution here?

 

[steph_swords]

you can factor out an A from the first equation and be left with A(B) = 1, am I on the right track?

 

[hotvette]

Yes. And what does that tell you about B?

 

[steph_swords]

that it's equal to the inverse of A.

 

[hotvette]

Yep. Now you should be in a position to answer the original questions.

 

[steph_swords]

okay I want to 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?

 

[Ray Vickson]

okay I want to 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

 

[steph_swords]

yeah that would work, I guess it's one of those times the simplest answer is the right answer. thanks for all your help.

 

[HallsofIvy]

From $A^3 - 2A^2 + 3A - I = 0$ we can get $A^3- 2A^2+ 3A= I$ and then $A(A^2- 2A+ 3)= I$. Do you see the point?

 

[Curious3141]

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 $B^{-1}$.