Everyone has solved some version of a linear system in either high school or college mathematics. If you've been keeping up with some of my other posts on algebra, you know that I'm about to either take something familiar away, or twist it into a different form. This time is no different; we're going to change the field we operate over, and solve a basic linear system in a Galois field called GF(4).

We already know how to compute the determinant of a $2\times 2$ matrix, so we're basically done. Now, I used very particular notation. I wrote $\text{det}(A)^{-1}$ instead of $\frac{1}{\text{det}(A)}$ on purpose. I wanted to indicate that it is the multiplicative inverse of the number that is $\text{det}(A)$. We have to be careful when discussing multiplication and division, because division as an operation doesn't always exist. This will be important when I twist around addition and multiplication soon.

Using Cramer's rule, we can solve our equation this way. The multiplicative inverse of 3 is the fraction 1/3, because multiplying these two together gives us 1, the multiplicative identity of the real numbers.

$$x = \frac{1}{3}\cdot \text{det}(A_{x}) = \frac{1}{3}\cdot 3 = 1$$ $$y = \frac{1}{3}\cdot\text{det}(A_{y}) = \frac{1}{3}\cdot 3 = 1$$ And we have our answer again, just obtained in a different way.A

- $(F,+)$ is an abelian group (a group where a+b = b+a.)
- $F$ is closed under multiplication.(The product of two elements doesn't leave the set F)
- The nonzero elements of $F$ form an abelian group under $\cdot$. (0 is the identity element for the operation +)
- We get the distributive law: $(a+b)\cdot c = (a\cdot c) + (b\cdot c)$

We just added an operation and some requirements to make sure nothing too weird happens. We can create addition and multiplication tables just like we did here. We're going to take a look at an example of a very specific field called a**Galois field.**This is simply a finite field with $q$ elements, if it exists.[note]Fun fact: Galois fields only exist if the number of elements is a prime or a power of a prime. So there is no Galois field with 6 elements. We denote a Galois field with $q$ elements $\text{GF}(q)$.

- 0 is clearly its own additive inverse. 0+0 = 0
- For 1, which element added to 1 returns 0? Looking at the addition table, we see that 1+1 = 0. Thus, 1 is its own additive inverse. Another way to write it is that -1 = 1, where $-x$ denotes the additive inverse of the element $x$ and
**not**the number $-1$ multiplied by $x$ - Continuing this exercise, we find that each element of GF(4) is its own additive inverse.

- 0 has no multiplicative inverse. Is GF(4) broken? No. Recall part (3) of the field definition above. The elements
*excluding*0 must form an abelian group under multiplication. That means it's totally ok for 0 times everything to return 0, and for 0 to have no multiplicative inverse. We did this on purpose. If we didn't have this definition, the real numbers wouldn't be a field either. When generalizing mathematics, we don't want to break what we already have. - 1 is its own multiplicative inverse. Makes sense. It's the multiplicative identity.
- The multiplicative inverse of 2 is 3 from the table, so the multiplicative inverse of 3 is 2. $2\cdot 3 = 3\cdot 2 = 1$.