Finding A: Solving the Matrix Equation

[yoleven]

Homework Statement

Find "a" when A is a square matrix satisfying (A+I)(A-I)=I and (A¹⁰¹)^-1=2^axA

I is the identity matrix.

The Attempt at a Solution

I'm trying to find A. I didn't know where to begin, so I picked A to be all zeroes and plugged it in the equation. It didn't work...
I tried A =
-1 -1
-1 -1
I ended up with
1 2
2 1

I want
1 0
0 1

Can some one give me a hint please.

 

[Dick]

The point isn't to find the matrix A, it's to find the number 'a'. Expand the first equation and learn something about A^2. Multiply both sides of the second equation by A^(101). Hmm?

 

[yoleven]

Okay.
A²=I+I²

A=$$\sqrt{I+I^2}$$

2^aA¹⁰²=1

I don't see a substitution that will help there.

I didn't know I could expand the first equation.

I'm not sure if I even multiplied through by A¹⁰¹ correctly.

 

[Dick]

You've got A^2=2I since I^2=I. Don't bother with the sqrt, you don't need to find A and you can't do it that way anyway. A^(102)=(A^2)^51. Now do you see it?

 

[yoleven]

A²=I

2^a(A²)⁵¹=1

2^a(I)⁵¹=1

2^a=(I)^-51

a ln 2=-51 ln I

a= -51(ln I/ ln 2)

a= -51 ln (I-2)

Is that close?

 

[Mark44]

A²=I

(above) No, A² = 2I.

2^a(A²)⁵¹=1

Should be I, not 1, on the right side.

2^a(I)⁵¹=1

2^a=(I)^-51

a ln 2=-51 ln I

a= -51(ln I/ ln 2)

a= -51 ln (I-2)

Is that close?

 

[HallsofIvy]

Also $(A^2)^{51}= A^{102}$, not $A^{101}$

And note that you want $A^{-101}$.

Knowing that $A^2= 2I$,what is $A^{-2}$?

It also helps to know that 101= 2(50)+ 1.

 

[yoleven]

I don't see how it is supposed to be an I on the right side instead of 1.
If I have (A¹⁰¹)^-1 that's just 1/A¹⁰¹.
If I multiply through by A¹⁰¹ then don't I have a 1 on the right side?

I tried this;

(A¹⁰¹)^-1=2^aA

$$\frac{1}{(A^2)^5^0}$$=2^a

2I^-50=2^a

 

[Jasso]

I don't see how it is supposed to be an I on the right side instead of 1.

Because both A and I are matrices. You mismatch the elements if you set it equal to 1.

If I have (A¹⁰¹)^-1 that's just 1/A¹⁰¹.
If I multiply through by A¹⁰¹ then don't I have a 1 on the right side?

No, because A*A^-1 = I, not 1.

Remember, as was pointed out before, A² = 2*I and A¹⁰¹=(A²)⁵⁰*A

 

[Mark44]

I don't see how it is supposed to be an I on the right side instead of 1.
If I have (A¹⁰¹)^-1 that's just 1/A¹⁰¹.

No, no, no! Matrix division is not defined!

If I multiply through by A¹⁰¹ then don't I have a 1 on the right side?

I tried this;

(A¹⁰¹)^-1=2^aA

Multiply both sides of the equation above by A¹⁰¹.
What is A¹⁰¹(A¹⁰¹)^-1?
Edit: Moved a right parenthesis.
What is A¹⁰¹2^aA?
What can you replace A² with?

$$\frac{1}{(A^2)^5^0}$$=2^a

2I^-50=2^a

As already noted, you can't divide by a matrix.

 

[HallsofIvy]

You can't define matrix "division" as "multiply by $A^{-1}$" for two reasons: 1) Many matrices do not have inverses.

2) If A does have an inverse, multiplying on left or right will typically give different results.

 

[yoleven]

Okay, I got it.

(A+I)(A-I)=I

A²-I²=I

A²=2I

(A²)⁵¹=(2I)⁵¹

A¹⁰²=2⁵¹I⁵¹

since I⁵¹=I

(A¹⁰²)^-1=(2⁵¹I)^-1

A^-102=2^-51I

A^-101xA^-1=2^-51I

A^-101=2^-51IA

A^-101=2^-51A

since originally, A^-101=2^axA

a must equal -51

thanks for all of your input.