Show that det(I-xy'T)= 1-y'T x

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Need Help Please!

I am very much in need of your help. I have a question saying:

Let x,y E R. Show that det(I-xy'T)= 1-y'T x

y'T is transpose of y and I is identity matrix.

Actually I don't know how to solve something like det( A-B). What am I going to do when there is addition or subtraction in the determinant.

Thanks for your response.
 
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Well for some square matrix M, you know how to calculate \det(M), right? And for two nxn square matrices A and B, A-B is an nxn matrix C, true again? Then calculating \det(A-B) is no different than calculating \det(C).
 
So is this so simple wov thank you very much I think I can handle the rest.
 
In future it might be a good idea not to say things like
"Let x,y E R" and then assert that they are matrices!
 
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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