matrix problem

[yoleven]

1. The problem statement, all variables and given/known data
Find "a" when A is a square matrix satisfying (A+I)(A-I)=I and (A101)-1=2axA

I is the identity matrix.



3. 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}

2aA102=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

2a(A2)51=1

2a(I)51=1

2a=(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.


2a(A2)51=1


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


2a(I)51=1

2a=(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 thats 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=2aA

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

2I-50=2a

[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 thats just 1/A101.
If I multiply through by A101 then don't I have a 1 on the right side?

No, becuase 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 thats 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=2aA

Multiply both sides of the equation above by A101.
What is A101(A101)-1?
Edit: Moved a right parenthesis.
What is A1012aA?
What can you replace A2 with?


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

2I-50=2a
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=2axA

a must equal -51

thanks for all of your input.