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Over $GF(4)$, received codeword is $z = 2?3$. Generator matrix is: $$ \left(\begin{array}{ccc}1&0&1\\0&1&3\end{array} \right) $$

From options 203, 213 and 223, what was the sent codeword?

I tried multiplying vector $u = (a\;b)$ by the generator matrix which would result in $c$ = (x y z) codeword (either 203, 213 or 223), but I got a failure on each try.

I know how to define actual sent codeword using parity check matrix, but I have no idea how to do it with generator matrix only.

Note: key says the correct answer is 223.

Lois2B
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  • Presumably you work with a slightly non-standard presentation of $GF(4)={0,1,2,3}$, where addition is bitwise XOR (so for example $1+1=0$) and multiplication is determined by distributivity and the rule $2\cdot2=3$. This is common enough in programming (that XOR is particularly efficient), but makes people with background only in algebra wince. – Jyrki Lahtonen Jun 10 '20 at 12:59
  • Anyway, you are looking for two elements $u,v\in GF(4)$ such that $$u(1,0,1)+v(0,1,3)=(2,?,3).$$ The first component on the left hand side is $u$, so by comparing we can deduce that $u=2$. Then the last component leads to the equation $$2+3v=3,$$ giving $3v=1$ and hence $v=2$. Therefore $$u(1,0,1)+v(0,1,3)=(2,0,2)+(0,2,1)=(2,2,3).$$ – Jyrki Lahtonen Jun 10 '20 at 13:04
  • See here for addition and multiplication tables of $GF(4)$. Your $2$ is the element denoted $B$ there, and your $3$ is renamed to $D$. – Jyrki Lahtonen Jun 10 '20 at 13:07

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