Let A be an invertible matrix. How to prove $\det (A^{-1}) = \frac{1}{\det A}$ without using the property that $\det (AA^{-1})=\det (A) \det(A^{-1})$? Thanks. I think $\det (A^{-1}) = \frac{1}{\det A}$ holds even without this property, given A is invertible.
I'm seeking for a proof of the following:     Let $A$ be an invertible matrix. Then the determinant of $A^{-1}$ equals:   $$\left|A^{-1}\right|=|A|^{-1} $$   I don't know where to begin the proof. Any suggestions?
Beamer: how to justify the whole text? How long does it take for badges to be awarded? How are they generated? Do badges get calculated in real time, or do they get calculated by an external service? How long does it take for a badge to be awarded, once its requirements are met?  Does Stack Overflow use any external services to do things, like maintenance, e-mailing or badge calculation? Crop a big picture into several small size pictures Is there a a way with GIMP editor to crop a big picture into several small equal sized pictures? So when all those small pictures connected in order would make a big picture. why is nautilus slow? I'm wondering why Nautilus is very slow when opening a directory containing lots of files. My /usr/lib dir for example has 1900 files and it takes approximately 5+ seconds to show everything. It has been like this since I installed Ubuntu few months ago and it's really quite annoying sometimes. 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However that question stipulated the answer could not be AutoKey. Given  $$ y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1. $$ Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ?  I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck. Proof of a formula for the power set of a union of two sets I need to prove that:  $2^{A\cup B} =\left \{ X\cup Y \mid X\in 2^{A}, Y\in2^{B} \right \}$  Note that $2^{A} = P(A)$ (The power set of A).  While I can prove the Right - Left direction fairly easily, It seems very hard for me to prove the Left - Right direction. Can I get some help with that?  How do I prove that $t\in2^{A\cup B}$ makes $t\in \left \{ X\cup Y \mid X\in 2^{A}, Y\in2^{B} \right \}$? Why are papers without code but with results accepted? 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Find the limit $$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$ I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks!  The cells of $T^2$ are $e^0, e^1_a, e^1_b, e^2$  The chain groups are  $$ C_0(T^2) = \{ k e^0 |&#160;k \in \mathbb{Z} \} = \mathbb{Z}$$  $$ C_1(T^2) = \{&#160;k_1 e^1_a + k_2 e^1_b |&#160;k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}$$  $$ C_2(T^2) = \mathbb{Z}$$  $$ C_k(T^2) = \{ 0 \} , k &gt; 2$$  Now the homology groups:  $$ H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}$$  where $im \partial_1 = 0$ because there is no chain in $C_1(T^2)$ whose boundary is a zero-chain in $C_0(T^2)$. (Is this reasoning correct?)  $$ H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}$$  where $\ker \partial_1 =  \mathbb{Z} \oplus \mathbb{Z} $, i.e. again everything maps to zero because there is no element in $C_1$ whose boundary maps to an element in $C_0$.  $im \partial_2 = 0$ again because there is no element in $C_2$ whose boundary is an element of $C_1$.  I don't want to use Hurewicz to do $H_1$.  $$ H_2(T^2) = \mathbb{Z}$$ using similar arguments as above.  Thanks for your help.  Edit  I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them... 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