$$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]}$$$

$$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]}$$$을(를) 구하시오.

먼저, 행렬을 대각화하세요(단계는 matrix diagonalization calculator를 참조하세요).

행렬이 대각화 가능하지 않으므로, 이를 대각행렬 $$$D = \left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]$$$와 멱영 행렬 $$$N = \left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]$$$의 합으로 나타내시오.

$$$N^{2} = \left[\begin{array}{cc}0 & 0\\0 & 0\end{array}\right]$$$라는 점에 유의하십시오.

이는 $$$e^{N} = I + N$$$임을 의미하며, 즉 $$$e^{\left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]} = \left[\begin{array}{cc}1 & 0\\0 & 1\end{array}\right] + \left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right] = \left[\begin{array}{cc}1 & - t\\0 & 1\end{array}\right]$$$입니다.

대각 행렬의 지수는 대각 원소에 지수 함수를 적용한 행렬이다: $$$e^{\left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]} = \left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]$$$

이제, $$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]} = e^{\left[\begin{array}{cc}t & 0\\0 & t\end{array}\right] + \left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]} = e^{\left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]}\cdot e^{\left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]} = \left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]\cdot \left[\begin{array}{cc}1 & - t\\0 & 1\end{array}\right].$$$

마지막으로, 행렬을 곱하십시오:

$$$\left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]\cdot \left[\begin{array}{cc}1 & - t\\0 & 1\end{array}\right] = \left[\begin{array}{cc}e^{t} & - t e^{t}\\0 & e^{t}\end{array}\right]$$$ (단계는 행렬 곱셈 계산기를 참조하세요).

$$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]} = \left[\begin{array}{cc}e^{t} & - t e^{t}\\0 & e^{t}\end{array}\right]$$$A