Question:
Let set G = { matrix [{a a},{a a}] such that a is real but not 0 } represent the set of 2x2 matrices with same elements of the reals excluding a = 0, show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinants.
Thoughts:
I can show closure, the existence of a unique identity and associativity; but the inverse has me lost. How can a 2x2 singular matrix with identical elements from the reals (excluding zero) be a group under matrix multiplication when no inverse exist that I can see (though the book asserts otherwise)?
Problem is from Contemporary Abstract Algebra (Gallian 8th Ed) Ch2 Problem 52.
$\endgroup$ 12 Answers
$\begingroup$Hint: for arbitrary $a \in \mathbb{R} \setminus \{0\}$, we're after a solution to the equation
$$\begin{bmatrix} a & a \\ a & a \\ \end{bmatrix} \begin{bmatrix} b & b \\ b & b \\ \end{bmatrix} = \mathrm{id}_G = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \\ \end{bmatrix}$$
$\endgroup$ 1 $\begingroup$A (adjA)= |A| I
here adj A = [a -a a -a ] is not in G. so
A^-1 = adjA/|A| doesnot mean in G
