Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to verify that the set

$\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$ where $a, b, c, d \in\mathbb{R}$, $ad - bc = 1$

is a group under $2 \times 2$ matrix multiplication

Please if someone can guide and help....

share|improve this question

closed as off-topic by Najib Idrissi, Old John, Daniel Fischer, amWhy, Mathmo123 Jul 31 at 11:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, Old John, Daniel Fischer, amWhy, Mathmo123
If this question can be reworded to fit the rules in the help center, please edit the question.

What have you tried? –  Tobias Kildetoft Jan 26 '13 at 2:54
homework should not be used as a standalone tag; see tag-wiki and meta. –  Martin Sleziak Jan 26 '13 at 2:55
If you're curious for more interesting properties of this group, it is called the special linear group of order 2, denoted $SL(2, \Bbb{R})$ –  Sam DeHority Jan 26 '13 at 3:16
@DenishSen A reminder: you can "accept" an answer (one per question) by clicking the check mark next to it. Your "accepted" answer is the one that you think best answered your question. This gives 15pts of reputation to the user whose answer you accepted, and ~2pt of reputation to yourself. –  anorton Jan 27 '13 at 22:04

2 Answers 2

This is what's called a "definition chase." You need to verify four things: that

  • The product of any two such matrices is also of this kind, that is, that the product of two matrices of determinant $1$ has determinant $1$;
  • Matrix multiplication is associative;
  • Any such matrix has an inverse of this type, that is, that the inverse of a matrix of determinant $1$ has determinant $1$;
  • There is an identity matrix with determinant $1$.

Really all you have to do is translate each of these three statements into symbols and verify that it's true.

share|improve this answer
Thank you Sir...for your guidance. –  Denish Sen Jan 27 '13 at 21:08

Recall that there are four requirements to be a group:

  1. The operation must be associative. That is $(a\circ b) \circ c = a\circ (b\circ c)$.
  2. The group must be closed. That is, $a\circ b$ is also satisfies the condition.
  3. There must exist an identity element, $e$, in the set, such that $a \circ e = a$.
  4. There must exist an inverse element, $a^{-1}$, in the set, such that $a \circ a^{-1} = e$.

Requirement 1:

Matrix multiplication is associative. See number 13 on this page: http://mathworld.wolfram.com/MatrixMultiplication.html

Requirement 2:

This can be verified.

Requirement 3:

This one is simple: use the $2\times 2$ identity matrix, $I$: $$I = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right]$$ It can be shown that $A\times I = A$.

Requirement 4:

The inverse matrix, $A^{-1}$ satisfies the equation $AA^{-1}=I$, by definition. The inverse matrix of A exists iff $\det A \not = 0$. The determinant of the matrix $A$ given below is $ad-bc$. $$A = \left[\begin{array}{cc} a & b \\ c & d\end{array}\right]$$ By the problem statement, we know the determinant is equal to one. Thus, the inverse matrix exists.

share|improve this answer
Are you sure that's the identity matrix? –  Sam DeHority Jan 26 '13 at 3:15
@DoctorBatmanGold Oof. Thanks. (corrected) It's been a long day. –  anorton Jan 26 '13 at 3:16
No problem. I was going to edit it myself but it was too few characters. –  Sam DeHority Jan 26 '13 at 3:19
@Downvoter: Just curious, why? There is nothing factually wrong with my answer (now), and it provides some guidance (but not all) for a homework problem. –  anorton Jan 26 '13 at 3:21
@anorton I didn't downvote but in requirement 4 it should say $\det A \ne 0$ not $\det I \ne 0$ –  Rustyn Jan 26 '13 at 4:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.