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 Post subject: Matrix with modulo
PostPosted: Tue, 8 Nov 2011 08:39:32 UTC 
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Joined: Sat, 20 Aug 2011 12:38:23 UTC
Posts: 39
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using
the (adapted) Gauss-Jordan Algorithm:

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all


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 Post subject: Re: Matrix with modulo
PostPosted: Tue, 8 Nov 2011 08:43:41 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13974
Location: Austin, TX
aceminer wrote:
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using
the (adapted) Gauss-Jordan Algorithm:

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all


You're trying to find a matrix B such that BA=AB=I

Hint:

\begin{pmatrix} 5 & 3 & 16 \\ 10 & 18 & 8 \\ 3 & 16 & 9\end{pmatrix}\equiv \begin{pmatrix} 5 & 3 & -3 \\ -9 & -1 & 8 \\ 3 & -3 & 9\end{pmatrix}\mod 19 will simplify the number of calculations you need to do.

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 Post subject: Re: Matrix with modulo
PostPosted: Tue, 8 Nov 2011 09:04:03 UTC 
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Joined: Sat, 20 Aug 2011 12:38:23 UTC
Posts: 39
Shadow wrote:
aceminer wrote:
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using
the (adapted) Gauss-Jordan Algorithm:

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all


You're trying to find a matrix B such that BA=AB=I

Hint:

\begin{pmatrix} 5 & 3 & 16 \\ 10 & 18 & 8 \\ 3 & 16 & 9\end{pmatrix}\equiv \begin{pmatrix} 5 & 3 & -3 \\ -9 & -1 & 8 \\ 3 & -3 & 9\end{pmatrix}\mod 19 will simplify the number of calculations you need to do.



But how does the mod 19 come into play.
i am given the mapping of a to a^- 1 mod 19 values.

I am not sure how to even start. Do i reduce the matrix to its inverse form to find A^-1 first? but the inverse encompasses a such that
[(1/3)(A)]^-1.
Not too sure if 1/3 will play a part in this.


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 Post subject: Re: Matrix with modulo
PostPosted: Tue, 8 Nov 2011 09:23:38 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13974
Location: Austin, TX
aceminer wrote:
Shadow wrote:
aceminer wrote:
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using
the (adapted) Gauss-Jordan Algorithm:

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all


You're trying to find a matrix B such that BA=AB=I

Hint:

\begin{pmatrix} 5 & 3 & 16 \\ 10 & 18 & 8 \\ 3 & 16 & 9\end{pmatrix}\equiv \begin{pmatrix} 5 & 3 & -3 \\ -9 & -1 & 8 \\ 3 & -3 & 9\end{pmatrix}\mod 19 will simplify the number of calculations you need to do.



But how does the mod 19 come into play.
i am given the mapping of a to a^- 1 mod 19 values.

I am not sure how to even start. Do i reduce the matrix to its inverse form to find A^-1 first? but the inverse encompasses a such that
[(1/3)(A)]^-1.
Not too sure if 1/3 will play a part in this.


The mod 19 is to tell you how to calculate inverses multiplicatively. Just apply the algorithm as usual with this new way of inverting stuff.

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 Post subject: Re: Matrix with modulo
PostPosted: Tue, 8 Nov 2011 09:54:25 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6782
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Topic moved from Foundations to Matrix Algebra.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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