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31 votes
31 votes

Consider the set $H$ of all $3 * 3$ matrices of the type $$\left( \begin{array}{ccc} a & f & e \\ 0 & b & d \\ 0 & 0 & c \end{array} \right)$$ where $a,b,c,d,e$ and $f$ are real numbers and $abc ≠ 0.$ Under the matrix multiplication operation, the set $H$ is:

  1. a group
  2. a monoid but not a group
  3. a semi group but not a monoid
  4. neither a group nor a semi group

4 Answers

Best answer
51 votes
51 votes

Given Information: Matrix is upper triangular. It's determinant is multiplication of principle diagonal elements. i.e., $abc.$
It is given that $abc \neq 0.$ So, Inverse for every such matrix exists.

Now this set is

  1. Closed - You can see after multiplication matrix is in same format and $|AB| = |A||B| \neq 0$ as $|A|,|B| \neq 0$
  2. Associative - Matrix multiplication is associative
  3. Existence of Identity - Identity Matrix is present
  4. Existence of Inverse - as determinant is non zero there exist inverse for every matrix

So, it is group.

Correct Answer: $A$

edited by
1 votes
1 votes

$abc\neq0$

The product of diagonal elements in a triangular matrix is the determinant.

=> determinant of such matrices $\neq0$

=> Matrices are non-singular

=> Matrices are invertible. -----------> #1

 

Closure holds.

Associativity holds. Matrix Chain Multiplication in Dynamic Programming is an example of this.

Identity holds. (The identity matrix)

Inverse holds. // From #1

Commutativity doesn't hold. As A.B $\neq$ B.A for matrices.

 

So, this is a group. Option A

0 votes
0 votes
A group
0 votes
0 votes

if any matrix is non singular means it has non zero determinant then it is

-closure

-Associativity

-Identity

-Inverse

so it is group 

Answer:

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