---
title: "NUS-MATH530 Some Matrix Manipulation"
source: https://www.jemoka.com/posts/kbhnus_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 \(a&rsquo;_1\) and \(a&rsquo;_2\). Given they are both inverses for \(a\), we have that:
\begin{equation} a * a&rsquo;_1 = 1 \end{equation}
as well as
\begin{equation} a * a&rsquo;_2 = 1 \end{equation}
Therefore, we have by the transitive property that:
\begin{equation} a * a&rsquo;_1 = a*a&rsquo;_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&rsquo;_1 = 1/a*a*a&rsquo;_2 \end{equation}
Therefore, that:
\begin{equation} a&rsquo;_1 = a&rsquo;_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 \(0&rsquo; \in F\).
Let&rsquo;s take some \(a \in F\). By definition of the additive identity, we have:
\begin{equation} 0 + a = a \end{equation}
We will apply \(0&rsquo;\) to both sides, we having that:
\begin{equation} 0&rsquo;(0+a) = 0&rsquo;a \end{equation}
Distributing \(0&rsquo;\) to both sides, we have:
\begin{equation} 1 + 0&rsquo;a = 0&rsquo;a \end{equation}
Given \(a,0&rsquo; \in F\), and multiplication is closed in \(F\) being a field, \(0&rsquo;a \in F\); applying \(-0&rsquo;a \in F\) the additive inverse of the result of multiplying together to both sides, we have that:
\begin{equation} 1 + 0&rsquo;a - 0&rsquo;a = 0&rsquo;a - 0&rsquo;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} &amp;3x + \frac{2}{3} = 3 \\ \Rightarrow\ &amp;3x = \frac{7}{3} \\ \Rightarrow\ &amp;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 &amp; 2 &amp; 1 \\ 2 &amp; 0 &amp; -1 \\ 1 &amp; -1 &amp; 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 &amp; b \\ c &amp;d \end{pmatrix} + \begin{pmatrix} e &amp; f \\ g &amp; h \end{pmatrix} = \begin{pmatrix} a+e &amp; b+f \\ c+g &amp; d+h \end{pmatrix} \end{equation}
Identity \begin{equation} \begin{pmatrix} a &amp; b \\ c &amp;d \end{pmatrix} + \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 0 \end{pmatrix} = \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix} \end{equation}
Inverse \begin{equation} \begin{pmatrix} a &amp; b \\ c &amp;d \end{pmatrix} + \begin{pmatrix} -a &amp; -b \\ -c &amp; -d \end{pmatrix} = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 0 \end{pmatrix} \end{equation}
Associative \begin{equation} \left ( \begin{pmatrix} x_1 &amp; x_2 \\ x_3 &amp; x_4 \end{pmatrix} + \begin{pmatrix} y_1 &amp; y_2 \\ y_3 &amp; y_4 \end{pmatrix} \right) + \begin{pmatrix} z_1 &amp; z_2 \\ z_3 &amp; z_4 \end{pmatrix} = \begin{pmatrix} (x_1+y_1)+z_1 &amp; (x_2+y_2)+z_2 \\ (x_3+y_3)+z_3 &amp; (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) &amp; x_2+(y_2+z_2) \\ x_3+(y_3+z_3) &amp; x_4+(y_4+z_4) \end{pmatrix} \end{equation}
And finally, this is equal to:
\begin{equation} \begin{pmatrix} x_1 &amp; x_2 \\ x_3 &amp; x_4 \end{pmatrix} + \left (\begin{pmatrix} y_1 &amp; y_2 \\ y_3 &amp; y_4 \end{pmatrix} + \begin{pmatrix} z_1 &amp; z_2 \\ z_3 &amp; 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 &amp; 0 \\ 0 &amp;1 \end{pmatrix} \end{equation}
is not invertable. In that, one cannot apply a matrix to this one to result in the multiplicative identity \(I_2\).
