Matrix Proof: A^t ~ B^t and A^-1 ~ B^-1 | Exam Time Tips

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If A~B prove :
a) A^t ~ B^t (Atranspose ~ Btranspose)
b) A^-1 ~ B^-1

I don't really know where to begin.. does the determinant have anything to do with it?

Exam time let's go people!
 
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dangish said:
If A~B prove :
a) A^t ~ B^t (Atranspose ~ Btranspose)
b) A^-1 ~ B^-1

I don't really know where to begin.. does the determinant have anything to do with it?

Exam time let's go people!

When you deleted most of the information in the template, you deleted the section on relevant formulas and definitions. One definition that is very important here is the definition of what it means when two matrices are similar.
 
When two matricies are similar, they have the same determinant.

However, because two matricies have the same determinant does not mean they are similar.

also know that detA^t=detA

Is this good stuff?
 
The definition that I'm familiar with has to do with matrix inverses.

If A and B are similar, then B = P-1AP for some invertible matrix P.

This is the way matrix similarity is usually defined.
 
Yes I know that one, I don't see where the transpose of a matrix comes into play there
 
Since A and B are similar (given) then B = P-1AP, for some invertible P.

If two matrices are equal, what can you say about their transposes? Also, when you have the transpose of a product, what does that equal?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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