Question

On \(R=\left\{\begin{pmatrix}a&b\\0&c\end{pmatrix}: a,b,c\in\mathbb{R}\right\}\) with the usual addition and multiplication of matrices, which of the following statements is true?

• A. R is a ring without zero-divisors.
• B. R is a ring with zero-divisors.
• C. R is a commutative ring.
• D. Every non-zero element in R has a multiplicative inverse.

Hint:

Look for a nonzero nilpotent element of \(R\), e.g. \(\begin{pmatrix}0&1\\0&0\end{pmatrix}\).

Answer 1

The Correct Option is B

Step 1: \(R\) is the set of \(2\times2\) upper triangular real matrices, which is closed under addition and multiplication, so it forms a ring.

Step 2: Check commutativity with \(A=\begin{pmatrix}1&0\\0&0\end{pmatrix}\) and \(B=\begin{pmatrix}0&1\\0&0\end{pmatrix}\). Then \(AB=\begin{pmatrix}0&1\\0&0\end{pmatrix}\) while \(BA=\begin{pmatrix}0&0\\0&0\end{pmatrix}\). Since \(AB\neq BA\), \(R\) is not commutative, so option (C) is false.

Step 3: Consider the nonzero element \(N=\begin{pmatrix}0&1\\0&0\end{pmatrix}\). Then \(N^2=\begin{pmatrix}0&0\\0&0\end{pmatrix}\). Since \(N\neq0\) but \(N\cdot N=0\), \(N\) is a nonzero zero-divisor, so \(R\) has zero-divisors and option (A) is false while option (B) holds.

Step 4: An invertible element can never be a zero-divisor, so the nilpotent element \(N\) has no multiplicative inverse. Hence not every nonzero element of \(R\) is invertible, so option (D) is false.

\[\boxed{\text{R is a ring with zero-divisors}}\]