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The set of all non-singular square matrices of same order with respect to matrix multiplication is
- a)quasi-group
- b)monoid
- c)group
- d)abelian group
Correct answer is option 'C'. Can you explain this answer?
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The set of all non-singular square matrices of same order with respect...
Explanation:
To determine whether the set of all non-singular square matrices of the same order form a group with respect to matrix multiplication, we need to check whether it satisfies the four group axioms:
1. Closure: The product of any two non-singular square matrices of the same order is also a non-singular square matrix of the same order. Therefore, the set is closed under matrix multiplication.
2. Associativity: Matrix multiplication is associative, which means that for any three matrices A, B, and C of the same order, (AB)C = A(BC). Since matrix multiplication is associative, the set satisfies the associativity property.
3. Identity element: The identity matrix I, which is a non-singular square matrix of the same order as any matrix in the set, serves as the identity element. For any matrix A in the set, AI = A and IA = A. Therefore, the set contains an identity element.
4. Inverse element: For every non-singular square matrix A in the set, there exists an inverse matrix A^(-1) such that AA^(-1) = A^(-1)A = I, where I is the identity matrix. The inverse of A is also a non-singular square matrix of the same order. Therefore, the set contains inverse elements for every matrix.
Since the set of all non-singular square matrices of the same order satisfies all four group axioms, it can be concluded that it forms a group with respect to matrix multiplication.
Hence, the correct answer is option 'C' - group.
To determine whether the set of all non-singular square matrices of the same order form a group with respect to matrix multiplication, we need to check whether it satisfies the four group axioms:
1. Closure: The product of any two non-singular square matrices of the same order is also a non-singular square matrix of the same order. Therefore, the set is closed under matrix multiplication.
2. Associativity: Matrix multiplication is associative, which means that for any three matrices A, B, and C of the same order, (AB)C = A(BC). Since matrix multiplication is associative, the set satisfies the associativity property.
3. Identity element: The identity matrix I, which is a non-singular square matrix of the same order as any matrix in the set, serves as the identity element. For any matrix A in the set, AI = A and IA = A. Therefore, the set contains an identity element.
4. Inverse element: For every non-singular square matrix A in the set, there exists an inverse matrix A^(-1) such that AA^(-1) = A^(-1)A = I, where I is the identity matrix. The inverse of A is also a non-singular square matrix of the same order. Therefore, the set contains inverse elements for every matrix.
Since the set of all non-singular square matrices of the same order satisfies all four group axioms, it can be concluded that it forms a group with respect to matrix multiplication.
Hence, the correct answer is option 'C' - group.
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Community Answer
The set of all non-singular square matrices of same order with respect...
To prove any set is a group with respect to certain operation we need to show
1- that operation is binary
Now the set should satisfy following properties with respect to the binary operation
2-Associativity
3-Existence of identity
4-Existence of inverse
and that's how we get "group"
now getting on the question
matrix multiplication of the same order will give the same order as a result therefore matrix multiplication is binary operation ( Binary operation always means closure)
now to prove other 3 properties
matrix multiplication is always associative in general
i.e. A.(B.C)=(A.B).C
for identity we know identity matrix is non- singular therefore it will belong to our set
now since our set contains square matrices which are non - singular therefore all of them are invertible
hence all the criteria of being a group is satisfied therefore our given set is group wrt to the Operation matrix multiplication
1- that operation is binary
Now the set should satisfy following properties with respect to the binary operation
2-Associativity
3-Existence of identity
4-Existence of inverse
and that's how we get "group"
now getting on the question
matrix multiplication of the same order will give the same order as a result therefore matrix multiplication is binary operation ( Binary operation always means closure)
now to prove other 3 properties
matrix multiplication is always associative in general
i.e. A.(B.C)=(A.B).C
for identity we know identity matrix is non- singular therefore it will belong to our set
now since our set contains square matrices which are non - singular therefore all of them are invertible
hence all the criteria of being a group is satisfied therefore our given set is group wrt to the Operation matrix multiplication
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The set of all non-singular square matrices of same order with respect to matrix multiplication isa)quasi-groupb)monoidc)groupd)abelian groupCorrect answer is option 'C'. Can you explain this answer?
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