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Exercise :

Solve the following system: $$ (S):\begin{cases} A+B+C=2^{n} & \\ A+j\;B+j^{2}\;C=(−1)^{n}\;j^{2n} & \\ A+j^{2}\;B+j\;C=(−1)^{n}\;j^{n} & \\ \end{cases} $$ Using the coefficients $1,j,j^{2}$ and $1+j+j^{2}=0$

Solution : \begin{aligned} A&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}j^{2n}+(-1)^{n}j^{n}\right) \\ &=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}2\cos\left( \dfrac{2n\pi}{3}\right)\right)\\ B&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}j^{2n+2}+(-1)^{n}j^{n+1}\right) \\ &=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}2\cos\left( \dfrac{(2n+1)\pi}{3}\right)\right)\\ C&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}j^{2n+1}+(-1)^{n}j^{n+2}\right) \\ &=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}2\cos\left( \dfrac{(2n-1)\pi}{3}\right)\right)\\ \end{aligned}

My question i can't get the same solution as the book : so here is my attempt using Gaussian elimination (row reduction )

$$\fbox{$\forall j \in [|2,n|]\quad L_j \longleftarrow L_j-\dfrac{a_{1,j}}{a_{1,1}}L_{1} $}$$

$$ S \iff \begin{cases} A+B+C=2^n \\ (j-1)B+(j^{2}-1)C=(-1)^{n}j^{2n}-2^n\\ (j^{2}-1)B+(j-1)C=(-1)^{n}j^{n}-2^n \end{cases} \iff \begin{cases} A+B+C=2^n \\ (j-1)B+(j^{2}-1)C=(-1)^{n}j^{2n}-2^n\\ -2C=(-1)^{n}j^{n}-2^n-\dfrac{1}{(j-1)}\left((-1)^{n}j^{2n}-2^n \right) \end{cases} $$

  • Could someone elaborate the calculation
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    You can add all three equations to eliminate $B$ and $C$. This immediately gives you $A$. – 2017-02-16

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Hint: In order to get $A$ add all equations note that the coefficients of B and C will be equal to zero. So you directly get $A$. Then Multiply the second equation with $j$ and subtract the second and third equation, don't forget to plug in $A$.

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    Better multiply the second by j² and third one by j then add to find B. Do the same with j and j² for C. – 2017-02-16