A $2 \times 2$ real matrix $N$ is given by $N = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix}$. The matrix exponential $e^{t N}$ is given by:

(A) $\begin{bmatrix} 1 & 3t \\ 0 & 1 \end{bmatrix}$

(B) $\begin{bmatrix} 1 & e^{3t} \\ 0 & 1 \end{bmatrix}$

(C) $\begin{bmatrix} e^t & 3t \\ 0 & e^t \end{bmatrix}$

(D) $\begin{bmatrix} 0 & 3e^t \\ 0 & 0 \end{bmatrix}$

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### 1. Strategy & Theoretical Principles By the Taylor series definition of the matrix exponential: $$e^{t N} = I + t N + \frac{t^2}{2!} N^2 + \frac{t^3}{3!} N^3 + \dots$$ If $N$ is a nilpotent matrix of index $k$ (i.e., $N^k = 0$), the infinite series terminates after $k-1$ terms. ### 2. Analytical Formulation & Calculation Compute $N^2$: $$N^2 = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$$ Since $N^2 = 0$, $N$ is nilpotent of degree 2, meaning all higher powers $N^k = 0$ for $k \ge 2$. Therefore: $$e^{t N} = I + t N = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + t \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 3t \\ 0 & 1 \end{bmatrix}$$ ### 3. Distractor Forensics - **Option A:** Correct matrix exponential. - **Option B:** Erroneously exponentiates the off-diagonal element ($e^{3t}$). - **Option C:** Erroneously exponentiates the diagonal elements. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Never exponentiate individual matrix entries element-by-element. - **Standard Reference:** Strang, G., *Linear Algebra and Its Applications*, Chapter 5.

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