Is the Inverse Calculation Correct for Matrix Polynomial Equation?

[MoreDrinks]

I suppose the title should be "Matrix polynomial T or F" but whatever.

Homework Statement

True or false: if A²-2A+I=0, then A^-1=2I²-I

The Attempt at a Solution

My thought:

A²-2A+I=0 becomes (A-I)(A-I)=0
so
A = I

The inverse of I is I.

So the second equation:

A^-1=2I²-I

becomes

I = 2I²-I

I²=I

2I-I = I

So the answer is True.

Where did I go wrong?

 

[Dick]

XX=0 doesn't mean X=0 for matrices. Try to find a counter example for your claim. Try A=[[1,1],[0,1]].

 

[MoreDrinks]

XX=0 doesn't mean X=0 for matrices. Try to find a counter example for your claim. Try A=[[1,1],[0,1]].

Gaaah, thanks. Although the TorF question specified, now that I think of it, a 4x4 matrix, making figuring it out for a fraction of a point even harder. Jesus, the entire lack of class or book examples is really screwing everyone in this course over.

One other, if you have a moment.

A = PBP^-1

and thus A = B and detA=detB

Most people put "false" thinking that because B was between P and P^-1, they did not always constitute I and thus both A=B and detA=detB would be wrong. But the answer was true.

Sorry if I'm breaking any rules, I just felt silly make an entirely new thread for this.

 

[Dick]

Gaaah, thanks. Although the TorF question specified, now that I think of it, a 4x4 matrix, making figuring it out for a fraction of a point even harder. Jesus, the entire lack of class or book examples is really screwing everyone in this course over.

One other, if you have a moment.

A = PBP^-1

and thus A = B and detA=detB

Most people put "false" thinking that because B was between P and P^-1, they did not always constitute I and thus both A=B and detA=detB would be wrong. But the answer was true.

Sorry if I'm breaking any rules, I just felt silly make an entirely new thread for this.

Not breaking any rules. A doesn't have to be equal to B. At all. But det(PBP^(-1))=det(P)*det(B)*det(P^(-1)). Now you can rearrange that to det(P)*det(P^(-1))*det(B). So det(A) does equal det(B). If you assume det(P) is nonzero. Otherwise even that's not even true.

 

[MoreDrinks]

Not breaking any rules. A doesn't have to be equal to B. At all. But det(PBP^(-1))=det(P)*det(B)*det(P^(-1)). Now you can rearrange that to det(P)*det(P^(-1))*det(B). So det(A) does equal det(B). If you assume det(P) is nonzero. Otherwise even that's not even true.

Thank you. So basically, we have half of a test question that is wrong and another half that was not covered in class notes, nor in the textbook we're using - we were never told and there were never any questions suggesting we could rearrange determinants like that.

Any recommendations on textbooks that DO cover these totally important topics?

 

[MoreDrinks]

I mean, they're just numbers, but showing that we can DO it would have been nice.

Again, it's late for me.

 

[Dick]

Thank you. So basically, we have half of a test question that is wrong and another half that was not covered in class notes, nor in the textbook we're using - we were never told and there were never any questions suggesting we could rearrange determinants like that.

Any recommendations on textbooks that DO cover these totally important topics?

I'm not really in teaching these days. So I can't really recommend one. But I can't imagine any of them don't mention det(AB)=det(A)det(B). Determinants are just numbers. You can rearrange them any way you anyway you can rearrange numbers.

 

[jbunniii]

I suppose the title should be "Matrix polynomial T or F" but whatever.

Homework Statement

True or false: if A²-2A+I=0, then A^-1=2I²-I

As noted, this is false. However, what can be said is that
$$0 = A^2 - 2A + I = A(A - 2I) + I$$
so
$$A(2I-A) = I$$
and therefore ##A^{-1} = 2I - A##.