Question:
The ring of all $2 \times 2$ matrices over reals is
The ring of all $2 \times 2$ matrices over reals is
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Matrix rings are the classic example of non-commutative rings with zero divisors.
Updated On: Jul 3, 2026
- an integral domain
- a field
- a skew field
- non-commutative
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The property of commutativity in a ring requires $a \cdot b = b \cdot a$ for all elements.
Step 2: Meaning
Matrix multiplication is generally order-dependent.
Step 3: Analysis
For $2 \times 2$ matrices $A$ and $B$, $AB$ is usually not equal to $BA$. For example, $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, while reversing them gives $\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$.
Step 4: Conclusion
Since $AB \ne BA$, the ring of $2 \times 2$ matrices is non-commutative. Final Answer: (D)
The property of commutativity in a ring requires $a \cdot b = b \cdot a$ for all elements.
Step 2: Meaning
Matrix multiplication is generally order-dependent.
Step 3: Analysis
For $2 \times 2$ matrices $A$ and $B$, $AB$ is usually not equal to $BA$. For example, $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, while reversing them gives $\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$.
Step 4: Conclusion
Since $AB \ne BA$, the ring of $2 \times 2$ matrices is non-commutative. Final Answer: (D)
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