Proving A-1 = 3I - A for a Square Matrix A Satisfying A2 - 3A + I = 0

[iamsmooth]

1. Show that if a square matrix A satisfies A² - 3A + I = 0, then A^-1 = 3I - A



2. A^-1A = I and A^-1A = I and more that I can't think of



3. 3A = A² + I

A = (A² + I)/3

?


This question is weird
Anyone know how to do it?

 

[iamsmooth]

I'm so stupid I got the template wrong. I think I figured it out:

Show that if a square matrix A satisfies A² - 3A + I = 0, then A^-1 = 3I - A

A² - 3A + I = 0
A² - 3A = -I
A(A - 3I) = -I
A - 3I = -I/A

I/A = A^-1, therefore:

A^-1 = 3I - A

 

[lanedance]

i would take out the divide step, its not really defined for matricies, there is only multiplication by the inverse

so before that step, just muliply through by -1 and you're finished

 

[iamsmooth]

Oh you're right! Matrix multiplication doesn't actually exist >.>

Thanks for the input!

 

[Mark44]

Oh you're right! Matrix multiplication doesn't actually exist >.>

Matrix multiplication actually DOES exist. It's matrix division that doesn't exist.

 

[Mark44]

I'm so stupid I got the template wrong. I think I figured it out:

Show that if a square matrix A satisfies A² - 3A + I = 0, then A^-1 = 3I - A

A² - 3A + I = 0
A² - 3A = -I
A(A - 3I) = -I
A - 3I = -I/A

I/A = A^-1, therefore:

A^-1 = 3I - A

I buy it up to the step where you have A(A - 3I) = -I. After that, I'm not buying. First off, and as already noted, there is no matrix division, so -I/A doesn't make sense.

Second, you say that I/A = A^-1. What makes you think that A has an inverse? All you are given is that A is a square matrix. There is nothing said about A being invertible (i.e., having an inverse).

So how can you justify your last statement, that A^-1 = 3I - A?

Hint: A(A - 3I) = -I, or equivalently, A(3I - A) = I.

 

[iamsmooth]

I buy it up to the step where you have A(A - 3I) = -I. After that, I'm not buying. First off, and as already noted, there is no matrix division, so -I/A doesn't make sense.

Second, you say that I/A = A^-1. What makes you think that A has an inverse? All you are given is that A is a square matrix. There is nothing said about A being invertible (i.e., having an inverse).

So how can you justify your last statement, that A^-1 = 3I - A?

Hint: A(A - 3I) = -I, or equivalently, A(3I - A) = I.

Isn't it said that a matrix is invertible if the product is an identity matrix, whereas (3I - A) is the inverse of A, so you know it's invertible. Thus instead of dividing, you multiply both sides by the inverse of A which would make 3I - A = A^-1

Or wait is it AB = BA = I, so it has to commute? I'm not too sure how to prove that A is invertible in this case, if this isn't right. :(

 

[HallsofIvy]

What are A(3I- A) and (3I- A)A ?