If no one told you that M was a matrix you would write
T = ae<sup>Mt</sup> as the solution where a is the value of T when t=0.
It turns out that the equation works even if M is a matrix and T is a vector whose components are the temperatures if you write
<pre>T = e<sup>Mt</sup>a</pre> and it turns out that e<sup>Mt</sup> can be evaluated by the Taylor series just as if Mt were a real number.

<p>e<sup>M</sup> = Σ<sub>j≥0</sub>M<sup>j</sup>/j!
<p>e<sup>Mt</sup> is a square matrix the same size as M.
<p><a href=b.c>Some code</a> that <a href=results>agrees</a> with the naïve code.
(Routine <tt>expm</tt> exponentiates a matrix.)
<p>If x is negative the Taylor series for e<sup>x</sup> is expensive.
If exp(x) directly applies the Taylor series then 1.0/exp(-x) is faster than exp(x) for negative x.
Similarly if you suspect that most of the eigenvalues are negative 
inv(expm(−M)) is cheaper than expm(x) to compute.
<p>This hack avoids limitations on M such as symmetry or even lop-sidedness.
M can even be complex.