[VA] SRC#003- New Year 2019 Special
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01-16-2019, 09:50 AM
Post: #6
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RE: [VA] SRC#003- New Year 2019 Special
(01-16-2019 04:24 AM)Thomas Okken Wrote: A similar calculation can be done: \(\begin{bmatrix} a & u^4 & u^4 & u^4\\ 1 & a & u^4 & u^4\\ 1 & 1 & a & u^4\\ 1 & 1 & 1 & a \end{bmatrix} \begin{bmatrix} u^3\\ u^2\\ u\\ 1 \end{bmatrix}= \begin{bmatrix} au^3+u^6+u^5+u^4\\ u^3+au^2+u^5+u^4\\ u^3+u^2+au+u^4\\ u^3+u^2+u+a \end{bmatrix}= (u^3+u^2+u+a) \begin{bmatrix} u^3\\ u^2\\ u\\ 1 \end{bmatrix}\) Cheers Thomas |
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