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shamim2k8
 Post subject: Equation solve
PostPosted: Fri, 9 Dec 2011 10:14:33 UTC 
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How can I solve linear equations which have more variables than its equation for example how can I solve this

x+y+z=9

4x+8y+6z=23

Here three variable but two equation.Please tell me the procedure of solving such equation.

Thank you
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outermeasure
 Post subject: Re: Equation solve
PostPosted: Fri, 9 Dec 2011 10:50:15 UTC 
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Moved from Applied 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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outermeasure
 Post subject: Re: Equation solve
PostPosted: Fri, 9 Dec 2011 10:54:48 UTC 
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shamim2k8 wrote:
How can I solve linear equations which have more variables than its equation for example how can I solve this

x+y+z=9

4x+8y+6z=23

Here three variable but two equation.Please tell me the procedure of solving such equation.

Thank you


Start by finding a particular solution. Then find the kernel of
\begin{pmatrix} 1&1&1\\ 4&8&6 \end{pmatrix}

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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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Soroban
 Post subject: Re: Equation solve
PostPosted: Fri, 9 Dec 2011 14:48:42 UTC 
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Hello, shamim2k8!

Quote:
How can I solve linear equations which have more variables than equations?

. . \begin{array}{ccccc}x+y+z&=&9 & [1] \\ 4x+8y+6z&=&23 & [2] \end{array}

We will have an infinite number of solutions.
. . We must find a neat way to say this.


Eliminate a variable, say, x.

\begin{array}{cccccc}\text{Multiply [1] by 4:} & 4x + 4y + 4z &=& 36 \\ \text{Subtract it from [2]:} & 4x + 8y + 6z &=& 23 \end{array}

And we have: .4y + 2z \:=\:-13

. . Solve for $y\!:\;\;y \:=\:\frac{-13-2z}{4}


Substitute into [1]: .x + \dfrac{-13-2z}{4} + z \:=\:36

. . Solve for x\!:\;\;x \:=\:\dfrac{49-2z}{4}


We have: .\begin{bmatrix}x &=& \dfrac{49-2z}{4} \\ \\[-3mm] y &=& -\dfrac{13+2z}{4} \\ \\[-4mm] z &=& z \end{bmatrix}


On the right, replace z with a parameter t.

. . And we have: .\begin{Bmatrix}x &=& \frac{1}{4}(49-2t) \\ \\[-3mm] y &=& -\frac{1}{4}(13+2t) \\ z &=& t \end{Bmatrix}


We have a solution for every real number t.
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Shadow
 Post subject: Re: Equation solve
PostPosted: Fri, 9 Dec 2011 21:06:48 UTC 
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For completeness it should be stated that you are not guaranteed infinitely many solutions in general. For the system:

\\ x+y+z=0 \\ x+y+z=1 there are NO solutions, so just be careful of that.

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