Question #203305

Check if matrix

1 a b

[A ]= 0 1 c

0 0 1

Where a,b,c∈R

is an abelian group with respect to matrix multiplication.


1
Expert's answer
2021-06-13T16:08:23-0400

Let us check whether the set A={(1ab01c001):a,b,cR}A=\Big\{ \begin{pmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix}:a,b,c\in\mathbb R\Big\} is an Abelian group with respect to matrix multiplication. It is sufficiently to prove that AA is a subgroup of the general linear group GL3(R).GL_3(\mathbb R).


Let (1a1b101c1001),(1a2b201c2001)A.\begin{pmatrix}1 & a_1 & b_1\\ 0 & 1 & c_1\\ 0 & 0 & 1\end{pmatrix}, \begin{pmatrix}1 & a_2 & b_2\\ 0 & 1 & c_2\\ 0 & 0 & 1\end{pmatrix}\in A.


Taking into account that


(1a1b101c1001)(1a2b201c2001)=(1a2+a1b2+a1c2+b101c2+c1001)A\begin{pmatrix}1 & a_1 & b_1\\ 0 & 1 & c_1\\ 0 & 0 & 1\end{pmatrix}\cdot \begin{pmatrix}1 & a_2 & b_2\\ 0 & 1 & c_2\\ 0 & 0 & 1\end{pmatrix}= \begin{pmatrix}1 & a_2+a_1 & b_2+a_1c_2+b_1\\ 0 & 1 & c_2+c_1\\ 0 & 0 & 1\end{pmatrix}\in A and


(1ab01c001)1=(1aacb01c001)A,\begin{pmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix}^{-1}= \begin{pmatrix}1 & -a & ac-b\\ 0 & 1 & -c\\ 0 & 0 & 1\end{pmatrix}\in A, we conclude that AA is a subgroup of GL3(R),GL_3(\mathbb R),

that is AA is a group with respect to matrix multiplication.


Since (110010001)(100011001)=(111011001)\begin{pmatrix}1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\cdot \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}= \begin{pmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix} and


(100011001)(110010001)=(110011001),\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}\cdot \begin{pmatrix}1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}= \begin{pmatrix}1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}, we conclude that the group AA is not Abelian.


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