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I was computing the matrix exponential of

$$X = \begin{pmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{pmatrix}$$

by diagonalization. Wolfram Alpha and Symbolab both solve this problem correctly. After the diagonalization and taking the exponential of the diagonal matrix, the only thing left to do is to compute the product

$$e^X = Pe^DP^{-1} = \begin{pmatrix}1&-1&1\\ 1&0&-2\\ 1&1&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&e&0\\ 0&0&e^3\end{pmatrix}\begin{pmatrix}\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\\ -\frac{1}{2}&0&\frac{1}{2}\\ \frac{1}{6}&-\frac{1}{3}&\frac{1}{6}\end{pmatrix},$$

and this is where I ran into a problem. Symbolab computes this multiplication and obtains the value of $e^X$ that I was expecting (and that Wolfram Alpha and Symbolab both reported earlier), that is,

$$e^x = \begin{pmatrix}\frac{2+3e+e^3}{6}&\frac{-e^3+1}{3}&\frac{2-3e+e^3}{6}\\ \frac{-e^3+1}{3}&\frac{2e^3+1}{3}&\frac{-e^3+1}{3}\\ \frac{2-3e+e^3}{6}&\frac{-e^3+1}{3}&\frac{2+3e+e^3}{6}\end{pmatrix}.$$

But when I tried the same multiplication on Wolfram Alpha, all of the entries were off by $1/3$rd. I checked to make sure that I had typed everything in right, checked what Wolfram Alpha gave for the inverse to see if it matched, etc., and everything else seemed right.

Am I doing something wrong? Does Wolfram Alpha not know how to multiply matrices or am I just making a simple mistake somewhere?

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  • $\begingroup$ I get the constant term to be $({1 \over 3}, {1 \over 2}, -{1 \over 6})^T (1,1,1)$ so I believe the above is incorrect. $\endgroup$ – copper.hat Jun 20 '20 at 0:51
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    $\begingroup$ You wrote {{0,0,0},{0,e,0},{0,0,e^3}} in Wolfram Alpha but it should be {{1,0,0},{0,e,0},{0,0,e^3}}. The first entry is wrong. Here is the corrected result. $\endgroup$ – user1551 Jun 20 '20 at 0:52
  • $\begingroup$ @user1551 yup, that's it. Thanks. $\endgroup$ – Math2718 Jun 20 '20 at 0:54
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As user1551 pointed out in a comment, the answer was as simple as a missing $1$: If

$$D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}, \quad \textrm{then} \quad e^D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & e^3 \end{pmatrix}, \quad \textrm{not} \quad e^D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & e^3 \end{pmatrix}.$$

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