Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Inverse = Original (matrices)

Find all 2 x 2 matrices such that

The only way I could think of doing it is equating all matrix variables and solving simultaneously with my calculator. Is there an easier way to do this (cuz I sure couldn't do it by hand)!

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Inverse = Original (matrices)

I believe you can use the Rational Canonical Form for this.  Other than that, I see no other way besides what you mentioned.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Inverse = Original (matrices)

Aren't there infinite solutions to this? Consider:

a and b are free variables, and so you can get as many such matrices as you want.
(Also ±I are two more.)

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Inverse = Original (matrices)

Multiplying the first by d and the second by a and then equating → a[sup]2[/sup] = d[sup]2[/sup] ⇒ a = ±d.

Consider the following cases separately:

It turns out that case (ii) ⇒ a = d = 0 and becomes a special case of case (iv).

The complete solution:

Case (i) yields ±I, case (iii) yields the two middle sets and case (iv) yields the last set.

gayitri

Guest

Re: Inverse = Original (matrices)

Find all 2 x 2 matrices such that

The only way I could think of doing it is equating all matrix variables and solving simultaneously with my calculator. Is there an easier way to do this (cuz I sure couldn't do it by hand)!

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Inverse = Original (matrices)

Thanks for the help !