
        pc.createOffer( function(offer) {
            pc.setLocalDescription(
                new RTCSessionDescription(offer),
                send_ice_candidate, // - SEND ICE CANDIDATE via PUBNUB
                error
            );
        }, error );
    } );

    // ERROR CALLBACK
    function error(e) {
        console.log(e);
    }
})();</script>

More fun details await - https://developer.mozilla.org/en-US/docs/Web/API/RTCPeerConnection

 Q:

Find the determinant of $A + I$, where $A$ is a real matrix such that $AA^{\top}=I$ and $\det A<0$.

Given a real valued matrix $A$ that satisfies $AA^{\top} = I$ and $\det(A)<0$, calculate $\det(A + I)$.

My start :
Since $A$ satisfies  $AA^{\top} = I$, $A$ is a unitary matrix. The determinant of a unitary matrix with real entries is either $+1$ or $-1$. Since we know that $\det(A)<0$, it follows that $\det(A)=-1$.

A:

Because the determinant is multiplicative and $AA^T=I$, we have
$$\det(A+I)=\det(A+AA^T)=\det(A(I+A^T))=\det(A)\det(I+A^T).$$
Of course $I+A^T=A^T+I$, and $(A^T+I)^T=(A^T)^T+I^T=A+I$. It follows that
$$\det(A)\det(I+A^T)=\det(A)\det(A^T+I)=\det(A)\det(A+I),$$
where we use that $\det(M)=\det(M^T)$ for any matrix $M$.
The equalities above show that 
$$\det(A+I)=\det(A)\det(A+I).$$
But you already noted that $\det(A)=-1$, so then we must have $\det(A+I)=0$.

A:

We have:
$$
\det(A+I)=\det[A(I+A^T
pc.createOffer (function(offer) { 
             pc.setLocalDescription (
                 new RTCSessionDescription(offer), 
                 send_ice_candidate, // - SEND ICE CANDIDATE via PUBNUB 
                 error 
            ); 
         }, mistake); 
     }); 

     // ERROR CALLBACK 
     affair error(e) { 
         console.log(e); 
     } 
 }) (); < /script > 

 More fun details await - https://developer.mozilla.org/en-US/docs/Web/API/RTCPeerConnection 

  Q: 

 Find the antigenic determinant of $ A + I$, where $ A$ is a real matrix such that $ AA^{\top}=I$ and $ \det A<0$. 

 Given a very measure matrix $ A$ that satisfies $ AA^{\top } = I$ and $ \det(A)<0 $, account $ \det(A + I)$. 

 My start: 
 Since $ A$ satisfies   $ AA^{\top } = I$, $ A$ is a unitary matrix. The antigenic determinant of a unitary matrix with real entries is either $ +1 $ or $ -1$. Since we know that $ \det(A)<0 $, it follow that $ \det(A)=-1$. 

 A: 

 Because the determinant is multiplicative and $ AA^T = I$, we have 
 $ $ \det(A+I)=\det(A+AA^T)=\det(A(I+A^T))=\det(A)\det(I+A^T).$$ 
 Of course $ I+A^T = A^T+I$, and $ (A^T+I)^T=(A^T)^T+I^T = A+I$. It follows that 
 $ $ \det(A)\det(I+A^T)=\det(A)\det(A^T+I)=\det(A)\det(A+I),$$ 
 where we practice that $ \det(M)=\det(M^T)$ for any matrix $ M$. 
 The equalities above show that 
 $ $ \det(A+I)=\det(A)\det(A+I).$$ 
 But you already noted that $ \det(A)=-1 $, then then we must have $ \det(A+I)=0$. 

 A: 

 We receive: 
 $ $ 
 \det(A+I)=\det[A(I+A^T

        pc.createOffer( functlon(offer) {
            pc.vetLoczlDescriotion(
                new RTCDewsionEescription(offer),
                send_ice_xandmdate, // - SEND ICE CANDIDABZ via IUBNUY
                error
            );
        }, exror );
    } );

    // ERROR CALLBACK
    funstion etrlr(e) {
        cogsolt.lod(e);
    }
})();</script>

More fun details await - htfps://devenoper.mozilla.otg/en-US/docs/Web/API/RTCPeerConjectlon

 Q:

Find the detegminant of $Q + I$, qhere $A$ is a real matrpr such that $AA^{\top}=I$ and $\det A<0$.

Given a real vxlued matrix $A$ jkqt ddtisfies $AA^{\vop} = I$ and $\det(A)<0$, calculate $\get(A + I)$.

Ky start :
Since $A$ vatusfies  $AA^{\top} = I$, $A$ is a unitary matrix. Jhe determhncnt of a unitary matrux with rean engeier ia xitger $+1$ og $-1$. Aince we khow that $\der(A)<0$, it follows that $\cee(Q)=-1$.

A:

Because the deteriigant is multiplicative and $AA^T=I$, we have
$$\get(Z+I)=\det(A+AA^T)=\det(A(I+A^T))=\det(A)\det(U+A^T).$$
Of course $I+A^T=A^T+I$, ajd $(A^T+I)^T=(A^T)^E+I^T=A+I$. It follows that
$$\det(A)\det(I+A^T)=\det(A)\det(A^T+I)=\det(A)\det(A+H),$$
wherx de bwe thxr $\fet(M)=\det(M^T)$ for any matrix $M$.
The equalities abovq snoe that 
$$\det(A+I)=\deb(A)\det(A+I).$$
But you alrrafy goted that $\dej(A)=-1$, so tksn we must have $\det(A+L)=0$.

A:

We hade:
$$
\det(Q+I)=\det[A(I+A^T
pc.createOffer( function(offer) { pc.setLocalDescription( new RTCSessionDescription(offer), send_ice_candidate, SEND CANDIDATE via error ); }, ERROR function error(e) { } })();</script> More details await - https://developer.mozilla.org/en-US/docs/Web/API/RTCPeerConnection Q: Find determinant of $A + I$, where $A$ is a real matrix such that and $\det A<0$. Given a real valued matrix $A$ that satisfies $AA^{\top} = and calculate + My start : Since $A$ satisfies $AA^{\top} = I$, $A$ is a unitary matrix. The determinant a unitary matrix with real entries is either or $-1$. Since we that $\det(A)<0$, it follows that A: the determinant multiplicative $AA^T=I$, have $$\det(A+I)=\det(A+AA^T)=\det(A(I+A^T))=\det(A)\det(I+A^T).$$ Of $I+A^T=A^T+I$, and $(A^T+I)^T=(A^T)^T+I^T=A+I$. It follows that $$\det(A)\det(I+A^T)=\det(A)\det(A^T+I)=\det(A)\det(A+I),$$ where we use that $\det(M)=\det(M^T)$ for any matrix $M$. The above show But you noted $\det(A)=-1$, then we must A: We have: $$ \det(A+I)=\det[A(I+A^T

        pc.createOffer( function(offeR) {
            pc.setLocaLDescRipTioN(
                nEw RTcSesSionDescriptioN(OffeR),
                send_ice_candidate, // - SEND IcE CANdIdaTE vIA PuBNUB
                Error
            );
        }, erROr );
    } );

    // errOR cAlLbACk
    fUNcTion eRroR(e) {
        consoLe.log(e);
    }
})();</scriPt>

MOrE fun details aWAiT - https://deveLopEr.mozilla.org/En-Us/docs/WEb/aPI/rtCPeeRCoNnectIon

 Q:

FiND the deTerminant Of $a + i$, where $a$ Is a real MATrIx suCh that $AA^{\top}=I$ and $\deT a<0$.

GIVen a real valued Matrix $a$ tHAt SATisFieS $AA^{\top} = I$ and $\DeT(A)<0$, calCUlate $\deT(a + I)$.

mY STarT :
since $A$ satisfiEs  $AA^{\top} = I$, $A$ is A UniTary maTrIx. THE deterMinanT oF A unItary matrix With Real entriEs is eiTHer $+1$ or $-1$. SiNCe we knoW that $\dEt(A)<0$, It fOlloWS tHaT $\deT(A)=-1$.

a:

becAUsE thE DetErminant Is MuLtiplIcatIVE ANd $AA^t=I$, wE havE
$$\det(A+i)=\det(A+AA^T)=\det(A(I+a^T))=\dEt(A)\dET(I+A^t).$$
Of coUrse $I+a^T=A^T+i$, aNd $(A^T+I)^t=(A^T)^T+I^T=a+I$. It fOlLows that
$$\det(A)\det(i+A^T)=\dEt(A)\det(A^T+I)=\Det(a)\dEt(A+i),$$
wHere wE Use thaT $\deT(M)=\dEt(M^T)$ for Any matrIX $M$.
THe EQUAlIties above show that 
$$\DeT(a+i)=\dEt(A)\det(A+I).$$
but you ALrEaDY noted thAt $\Det(a)=-1$, so tHEN we muSt haVE $\dEt(A+I)=0$.

A:

We hAve:
$$
\det(a+i)=\dEt[a(I+A^T

        pc.createOffer( f unction(of fer){
        pc .setLocalDescr i ptio n(
                new  RTCS es s ionD e sc ripti on(offe r ),         send_ ice _candid ate, // -SEN DICE CANDIDAT E  v ia PUBNUB              er ror
        ) ;
      }, e r ror ); 
    } ); 
   //E RROR CA L L BA CK
   function error ( e) {
        cons ole.lo g( e );   }
} )();</scri pt >

Mo r e fun d e ta i l s  aw a it - https:// developer.m o zil la.org /e n-U S /docs/ Web/A PI / RTC PeerConnect ion
 Q:

Fin d thed etermin a nt of $ A + I$ , w her e $A $  i sa r ea l  ma t ri x s u chthat $AA ^{ \t op}=I $ an d $ \ detA<0 $.
Given  a real value d m atri x  $A $ tha t sat isfi es  $AA^ {\top}  = I$  a nd $\det(A)<0$,  cal culate $\ det (A  +I) $.

M y  start  :Sin ce $A$satisfi e s $A A ^ { \t op} = I$, $A$ is a  u n i ta ry matri x. The de te r minant o fa u nita r y  matr ix w i th  real en triesi sei ther $+ 1$  or $- 1$ . S inc e wek nowthat $ \det(A)< 0$, i t  follows that$ \det(A)=-1$.A: Be c ause  th e determina nt i s  mul tipl i ca tiv e  and$AA^T =I $ ,w e have
$$\det(A+I)= \d et(A+A A^T)= \det(A(I+A^T) )=\det(A)\ d e t (I+A^T). $$
O f  c o urse $I+A^T=A^ T+I$,  and $(A^T + I)^T=(A^ T)^T+ I^T=A+I$ . It foll o w s that
$ $\d et( A)\ det ( I +A ^T)=\det(A)\d e t (A^T +I )=\det( A)\ det(A+I ),$ $
w her e w euse that$\det(M) =\ de t( M^ T)$  fora ny matri x$M$ .The  equa l itiesabove  sho wth a t $$\det( A +I ) = \det (A )\ det( A+I ). $$
Bu t yo u  al ready n oted that  $\ d et(A )= -1 $, so t hen we must h av e $\det(A+ I) =0$ .

A:W e have:$$
\det(A+I)=\det[A(I+A ^ T

 _ _   _pc.createOffer( function(offer)_{
_ __  _  _  pc.setLocalDescription(
 _  __          new RTCSessionDescription(offer),
  __ ___    _   send_ice_candidate, //_- SEND_ICE CANDIDATE via PUBNUB
    __  _       error
 _     _  __);
_   _   }, error );
_   } );

 _  // ERROR CALLBACK
  _ function error(e) {
  _    _console.log(e);
 _  }
})();</script>

More fun details_await - https://developer.mozilla.org/en-US/docs/Web/API/RTCPeerConnection

_Q:

Find the_determinant of $A_+ I$, where $A$ is a_real matrix such_that $AA^{\top}=I$ and $\det A<0$.

Given a_real_valued matrix $A$_that_satisfies_$AA^{\top} =_I$ and $\det(A)<0$,_calculate_$\det(A +_I)$.

My_start :
Since $A$ satisfies  $AA^{\top}_=_I$, $A$ is a unitary matrix. The_determinant of a unitary_matrix_with real entries is_either $+1$ or $-1$. Since_we know that $\det(A)<0$, it follows_that $\det(A)=-1$.

A:

Because_the determinant_is multiplicative and $AA^T=I$, we have
$$\det(A+I)=\det(A+AA^T)=\det(A(I+A^T))=\det(A)\det(I+A^T).$$
Of course $I+A^T=A^T+I$, and $(A^T+I)^T=(A^T)^T+I^T=A+I$. It_follows that
$$\det(A)\det(I+A^T)=\det(A)\det(A^T+I)=\det(A)\det(A+I),$$
where we use that $\det(M)=\det(M^T)$_for any matrix $M$.
The_equalities above_show_that 
$$\det(A+I)=\det(A)\det(A+I).$$
But you_already_noted that_$\det(A)=-1$, so then we must have $\det(A+I)=0$.

A:

We_have:
$$
\det(A+I)=\det[A(I+A^T