NUS-MATH530 Some Matrix Manipulation

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.