Algebraic Structures | Question-3
Solution
Recall the definition of set \( B \):
The set \( B \) contains all \( 2 \times 2 \) matrices \( X \) that commute with matrix \( M \).
Definition of \( B \):
\[ B = \{ X \in A \mid MX = XM \} \]
This means for any \( X \in B \), the matrix equation \( MX = XM \) holds true.
Assume \( P \) and \( Q \) are elements of \( B \):
Given that \( P, Q \in B \), by definition we have:
Commutativity Conditions:
\[ MP = PM \quad \text{and} \quad MQ = QM \]
These are the defining properties of matrices in set \( B \).
Compute \( M(PQ) \) using matrix properties:
Associativity of Matrix Multiplication:
Using the associative property of matrix multiplication:
\[ M(PQ) = (MP)Q \]
Apply the commutativity condition for \( P \):
Substitute Commuting Relation for \( P \):
Since \( P \in B \), we know \( MP = PM \):
\[ M(PQ) = (MP)Q = (PM)Q \]
Apply associativity again:
Regroup Using Associativity:
Using the associative property of matrix multiplication again:
\[ (PM)Q = P(MQ) \]
Apply the commutativity condition for \( Q \):
Substitute Commuting Relation for \( Q \):
Since \( Q \in B \), we know \( MQ = QM \):
\[ P(MQ) = P(QM) \]
Apply associativity for the final step:
Final Associative Step:
Using the associative property once more:
\[ P(QM) = (PQ)M \]
Complete Chain of Equalities:
\[ M(PQ) = (MP)Q = (PM)Q = P(MQ) = P(QM) = (PQ)M \]
Conclude that \( PQ \in B \):
Satisfying the Definition:
Since \( M(PQ) = (PQ)M \), the matrix product \( PQ \) satisfies the defining condition for membership in \( B \).
Therefore, by definition of \( B \), we have:
\[ PQ \in B \]
Conclusion
Final Answer: ✅ PROVED
Proof Summary:
- Given \( P, Q \in B \), we have \( MP = PM \) and \( MQ = QM \)
- Using associativity: \( M(PQ) = (MP)Q \)
- Substituting \( MP = PM \): \( (MP)Q = (PM)Q \)
- Using associativity: \( (PM)Q = P(MQ) \)
- Substituting \( MQ = QM \): \( P(MQ) = P(QM) \)
- Using associativity: \( P(QM) = (PQ)M \)
- Thus: \( M(PQ) = (PQ)M \)
- Therefore: \( PQ \in B \)
Mathematical Significance: This proof shows that set \( B \) is closed under matrix multiplication, which means \( B \) forms a subalgebra of the algebra of \( 2 \times 2 \) matrices. Combined with closure under addition (from Question 2), this suggests \( B \) has a rich algebraic structure.
Alternative Approach: From Question 1, we know matrices in \( B \) have the form \( \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \). The product of two such matrices is:
\[ \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \begin{pmatrix} c & d \\ 0 & c \end{pmatrix} = \begin{pmatrix} ac & ad + bc \\ 0 & ac \end{pmatrix} \]
which also has the required form \( \begin{pmatrix} ac & ad+bc \\ 0 & ac \end{pmatrix} \), confirming \( PQ \in B \).
Verification: ✅ We have successfully proven that if \( P, Q \in B \), then \( P \cdot Q \in B \).