Proof: identity of a group is unique
Assume for contradiction that there exists two identities \(e_1\) and \(e_2\) which are identities of the group \(A\). Take also an \(a \in A\).
Given both \(e_1\) and \(e_2\) are identities, we have that:
\begin{equation} a * e_1 = a \end{equation}
as well as
\begin{equation} a * e_2 = a \end{equation}
Therefore, we have by the transitive property that:
\begin{equation} a * e_1 = a*e_2 \end{equation}
Because we are in a group, there exists a \(1/a\) the inverse of \(a\). Applying this inverse to the expression, we have that:
\begin{equation} 1/a*a * e_1 = 1/a*a*e_2 \end{equation}
Therefore, that:
\begin{equation} e_1 = e_2\ \blacksquare \end{equation}
Therefore, there cannot be two unique identities in a group.
Proof: inverse of an element in a group is unique
Take group \(A\) and element \(a\in A\), assume for contradiction that there exists two inverses of \(a\) named here \(a’_1\) and \(a’_2\). Given they are both inverses for \(a\), we have that:
\begin{equation} a * a’_1 = 1 \end{equation}
as well as
\begin{equation} a * a’_2 = 1 \end{equation}
Therefore, we have by the transitive property that:
\begin{equation} a * a’_1 = a*a’_2 \end{equation}
Because we are in a group, there exists a \(1/a\) the inverse of \(a\). Applying this inverse to the expression, we have that:
\begin{equation} 1/a*a * a’_1 = 1/a*a*a’_2 \end{equation}
Therefore, that:
\begin{equation} a’_1 = a’_2\ \blacksquare \end{equation}
Therefore, there cannot be two unique inverses for an element in group.
Proof: additive identity in field cannot have multiplicative inverse
For some field \(F\) take its additive identity \(0 \in F\). Assume for the sake of contradiction there exists a multiplicative inverse for \(0\) named \(0’ \in F\).
Let’s take some \(a \in F\). By definition of the additive identity, we have:
\begin{equation} 0 + a = a \end{equation}
We will apply \(0’\) to both sides, we having that:
\begin{equation} 0’(0+a) = 0’a \end{equation}
Distributing \(0’\) to both sides, we have:
\begin{equation} 1 + 0’a = 0’a \end{equation}
Given \(a,0’ \in F\), and multiplication is closed in \(F\) being a field, \(0’a \in F\); applying \(-0’a \in F\) the additive inverse of the result of multiplying together to both sides, we have that:
\begin{equation} 1 + 0’a - 0’a = 0’a - 0’a \end{equation}
And therefore:
\begin{equation} 1 = 0 \end{equation}
which is absurd, reaching the desired contradiction. \(\blacksquare\)
System
\begin{equation} \begin{cases} x + 2y + z = 0 \\ 2x + 0y - z = 1 \\ x - y + z = 2 \\ \end{cases} \end{equation}
We will subtract the top and bottom expressions to have that:
\begin{equation} 3y = -2 \end{equation}
And to get:
\begin{equation} y = \frac{-2}{3} \end{equation}
Manipulating the second expression, we have that:
\begin{equation} 2x -1 = z \end{equation}
Substituting this expression and \(y\) into the third expression, we have:
\begin{equation} x + \frac{2}{3} + 2x -1 = 2 \end{equation}
performing algebraic manipulations:
\begin{align} &3x + \frac{2}{3} = 3 \\ \Rightarrow\ &3x = \frac{7}{3} \\ \Rightarrow\ &x = \frac{7}{9} \end{align}
And finally:
\begin{equation} \frac{14}{9}-1 = z = \frac{5}{9} \end{equation}
Multiply
\begin{equation} \begin{pmatrix} 1 & 2 & 1 \\ 2 & 0 & -1 \\ 1 & -1 & 0\end{pmatrix} \begin{pmatrix} x \\ y\\ z \end{pmatrix} = \begin{pmatrix} x+2y+z \\ 2x-z \\ x-y \end{pmatrix} \end{equation}
The inner dimensions (column vs. row) of the matricies have to be the same for them to be multiplied; matrix multiplication is not commutative.
Proof: 2x2 Matrices with Real Entries form a Group Under Addition
Closure
\begin{equation} \begin{pmatrix} a & b \\ c &d \end{pmatrix} + \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} a+e & b+f \\ c+g & d+h \end{pmatrix} \end{equation}
Identity
\begin{equation} \begin{pmatrix} a & b \\ c &d \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{equation}
Inverse
\begin{equation} \begin{pmatrix} a & b \\ c &d \end{pmatrix} + \begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \end{equation}
Associative
\begin{equation} \left ( \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix} + \begin{pmatrix} y_1 & y_2 \\ y_3 & y_4 \end{pmatrix} \right) + \begin{pmatrix} z_1 & z_2 \\ z_3 & z_4 \end{pmatrix} = \begin{pmatrix} (x_1+y_1)+z_1 & (x_2+y_2)+z_2 \\ (x_3+y_3)+z_3 & (x_4+y_4)+z_4 \end{pmatrix} \end{equation}
which is equal, by associativity in \(\mathbb{F}\), as:
\begin{equation} \begin{pmatrix} x_1+(y_1+z_1) & x_2+(y_2+z_2) \\ x_3+(y_3+z_3) & x_4+(y_4+z_4) \end{pmatrix} \end{equation}
And finally, this is equal to:
\begin{equation} \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix} + \left (\begin{pmatrix} y_1 & y_2 \\ y_3 & y_4 \end{pmatrix} + \begin{pmatrix} z_1 & z_2 \\ z_3 & z_4 \end{pmatrix} \right) \end{equation}
We have therefore shown that 2x2 matricies form a group under addition.
Proof: 2x2 Matrices with Real Entries does not from a Group Under Multiplication
Inverse
The matrix
\begin{equation} \begin{pmatrix} 0 & 0 \\ 0 &1 \end{pmatrix} \end{equation}
is not invertable. In that, one cannot apply a matrix to this one to result in the multiplicative identity \(I_2\).