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# Matrix question q29 Re1 He p6 watch

1. given that n (is a member) of +integers, x is a member of REal numbers, and M= cosh^2 x cosh^2 x
sinh^2 x sinh^2 x (2x2 matrix)

use induction to show that M^n=M
I can show that M^2=m but not sure how to use the let n=k method!
2. (Original post by ogs)
given that n (is a member) of +integers, x is a member of REal numbers, and M= cosh^2 x cosh^2 x
sinh^2 x sinh^2 x (2x2 matrix)

use induction to show that M^n=M
I can show that M^2=m but not sure how to use the let n=k method!
You have to use "let n=k+1", throw into the equation and see if your result would be as expected.
3. M^3
= M^2 M
= M^2 . . . because M^2 = M
= M . . . again because M^2 = M

M^4
= M^3 M
= M^2 . . . because M^3 = M
= M . . . because M^2 = M

M^5
= M^4 M
= M^2 . . . because M^4 = M
= M . . . because M^2 = M
4. that won't work as a proof by induction though! as you have to consider a general value...
or will it? do you think that it would get the marks?
5. (Original post by ogs)
that won't work as a proof by induction though! as you have to consider a general value...
or will it? do you think that it would get the marks?
Since M^k.M = M.M = M^2 by induction hypothesis the inductive proof is equivalent to showing M^2 = M which can be done simply by multiply the matrices and seeing what you get (hopefully M!) remembering that cosh^2x - sinh^2x = 1
6. Here is a more formal presentation of the inductive step. Suppose that k >= 2 is an integer and M^k = M. Then

M^(k + 1)
= M^k M
= M^2 . . . because M^k = M
= M . . . proved in the first part of the question

So M^(k + 1) = M.

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Updated: June 28, 2004
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