Answer to Question #135130 in Abstract Algebra for GIRISH CHANDRA

Question

Give an example with justification of an element of M3 (Z) that is a unit but not the identity element.

Answer 1

$$Solution$$

$M3(Z)$ is a $3\ by\ 3$ matrix over $Z$

An element $M3(Z)=(a_{ij})3\ by\ 3$

$M3(Z) = \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{pmatrix}$

A $3\ by\ 3$ matrix over $Z$ can be defined as a unit if it has all its elements as zeros except for the main diagonal, which has only ones.

$M3(Z) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

We all know that an identity matrix is the one that satisfies

$A (+, \cdot) I_{nXn}=A$ Where A is any square matrix of order n

Now on additive identity of matrices, an identity is a matrix with all its elements as zeros.

$M3(Z)$ is a unit but in this case it cannot be an additive identity, for example;

Let A be a matrix $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}$

The additive identity

$A+M3(Z)= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}+\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}= \begin{pmatrix} 2 & 2 & 3 \\ 2 & 4 & 4 \\ 3 & 4 & 6 \end{pmatrix} \neq \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}$

But instead

$A+I_3= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}+\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}$

Hence proven that an element of $M3 (Z)$ that is a unit but not the identity element.