Let M_{2 times 2} (mathbb{Z}/mathbb{6Z}) be the set of 2 x 2 matrices with the entries in mathbb{Z}/mathbb{6Z} a) Can you find a matrix M_{2 times 2}

Mylo O'Moore

Mylo O'Moore

Answered question

2021-02-08

Let M2Γ—2(Z/6Z) be the set of 2 x 2 matrices with the entries in Z/6Z
a) Can you find a matrix M2Γ—2(Z/6Z) whose determinant is non-zero and yet is not invertible?
b) Does the set of invertible matrices in M2Γ—2(Z/6Z) form a group?

Answer & Explanation

Leonard Stokes

Leonard Stokes

Skilled2021-02-09Added 98 answers

Step 1
(a)
According to given condition in the first part (a), Yes, we can construct a matrix A=M2Γ—2(Z/6Z) whose determinant is non-zero and yet it is not invertible. Then the matrix A is:
A=[3βˆ’3βˆ’55]
Step 2
As the determinant of the matrix A is zero which is not invertible. But elements of matrix A from (Z/6Z). So, after applying the modulo 6, the matrix A becomes:
A=[3βˆ’3βˆ’55]
A=[3βˆ’3βˆ’55]
=[3315]βˆ’3β‰…3mod(6)βˆ’5β‰…1mod(6)
Step 3
This shows that, the determinant of matrix A is 12 which is non- zero but itself matrix A is not invertible.
|A|=|3315|
=(15βˆ’3)
=12
β‰ 0
Step 4
(b)
yes, the set of invertible matrices in M2Γ—2(Z/6Z) forms a group because it satisfies the following property of group:
1)For any two matrices A,B∈M2Γ—2(Z/6Z) then Aβ‹…B∈M2Γ—2(Z/6Z) then it satisfies closure property.
2) (Aβ‹…B)β‹…C=Aβ‹…(Bβ‹…C), for any three matrices A,B,C∈M2Γ—2(Z/6Z) then it satisfies the associative property.
Step 5
Identity: There exist an identity element κ“² should belong to the set of matrices M2Γ—2(Z/6Z) for all matrix A∈M2Γ—2(Z/6Z) so we have Aβ‹…I=Iβ‹…A=I then the identity element of the set of matrices is:
I=[1001]
Step 6
The existence of inverse: As the set of matrices M2Γ—2(Z/6Z) contains the invertible matrices. So that the inverse of every matrix exists. This implies that for every A∈M2Γ—2(Z/6Z) there exist Aβˆ’1∈M2Γ—2(Z/6Z) such that Aβ‹…Aβˆ’1=Aβˆ’1β‹…A=I then, Aβˆ’1∈M2Γ—2(Z/6Z)Β for allΒ A∈M2Γ—2(Z/6Z)
Step 7
This shows that the set of invertible matrices in M2Γ—2(Z/6Z) satisfies the all properties of a group. Hence, it forms a group.

Jeffrey Jordon

Jeffrey Jordon

Expert2022-01-30Added 2605 answers

Answer is given below (on video)

Jeffrey Jordon

Jeffrey Jordon

Expert2022-08-23Added 2605 answers

Answer is given below (on video)

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