I'm trying to fit this matrix within one column. But it is exceeding the limit and overlapping with the column beside of it.


$e^{At}$ = \[ \left[ \begin{array}{ccc}
3*e^{-t}-3*e^{-2t}+e^{-3t} & \dfrac{5}{2}*e^{-t}-4*e^{-2t}+\dfrac{3}{2}*e^{-3t} & \dfrac{1}{2}*e^{-t}-e^{-2t}+\dfrac{1}{2}*e^{-3t} \\\\
-3*e^{-t}+6*e^{-2t}-3*e^{-3t} & \dfrac{-5}{2}*e^{-t}+8*e^{-2t}-\dfrac{9}{2}*e^{-3t} & \dfrac{-1}{2}*e^{-t}+2*e^{-2t}-\dfrac{3}{2}*e^{-3t} \\\\
3*e^{-t}-12*e^{-2t}+9*e^{-3t} & \dfrac{5}{2}*e^{-t}-16*e^{-2t}+\dfrac{27}{2}*e^{-3t} & \dfrac{1}{2}*e^{-t}-4*e^{-2t}+\dfrac{9}{2})*e^{-3t} \end{array} \right]\]
  • Do you really need * to denote multiplication? – Sigur Sep 5 '15 at 18:01
  • No...not at all – pradeep vejju Sep 5 '15 at 18:02
  • You can delete all of them and also use only \begin{bmatrix}\end{bmatrix}. Also, use \\[1em] to change line instead of insert blank lines. – Sigur Sep 5 '15 at 18:04
  • 1
    Welcome to TeX.sx! – strpeter Sep 5 '15 at 18:05
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    You can always switch to a temporary smaller font {\small ... } or \small...\normalsize. – John Kormylo Sep 5 '15 at 18:33

I suggest you use the cuted package, and its strip environment, so as the equation spreads over two columns. I also reduced the size of the fraction coefficients using the \mfrac command (medium sized fractions), from the nccmath package:

\usepackage{mathtools, nccmath}


      3 e^{-t}-3 e^{-2t}+e^{-3t} & \mfrac{5}{2} e^{-t}-4 e^{-2t}+\mfrac{3}{2} e^{-3t} & \mfrac{1}{2} e^{-t}-e^{-2t}+\mfrac{1}{2} e^{-3t} \\[2ex]
      -3 e^{-t}+6 e^{-2t}-3 e^{-3t} & -\mfrac{5}{2} e^{-t}+8 e^{-2t}-\mfrac{9}{2} e^{-3t} & -\mfrac{1}{2} e^{-t}+2 e^{-2t}-\mfrac{3}{2} e^{-3t} \\[2ex]
      3 e^{-t}-12 e^{-2t}+9 e^{-3t} & \mfrac{5}{2} e^{-t}-16 e^{-2t}+\mfrac{27}{2} e^{-3t} & \mfrac{1}{2} e^{-t}-4 e^{-2t}+\mfrac{9}{2} e^{-3t}


enter image description here

| improve this answer | |

enter image description here


3e^{-t}-3e^{-2t}+e^{-3t} & \dfrac{5}{2}e^{-t}-4e^{-2t}+\dfrac{3}{2}e^{-3t} & \dfrac{1}{2}e^{-t}-e^{-2t}+\dfrac{1}{2}e^{-3t} \\[1em]
-3e^{-t}+6e^{-2t}-3e^{-3t} & \dfrac{-5}{2}e^{-t}+8e^{-2t}-\dfrac{9}{2}e^{-3t} & \dfrac{-1}{2}e^{-t}+2e^{-2t}-\dfrac{3}{2}e^{-3t} \\[1em]
3e^{-t}-12e^{-2t}+9e^{-3t} & \dfrac{5}{2}e^{-t}-16e^{-2t}+\dfrac{27}{2}e^{-3t} & \dfrac{1}{2}e^{-t}-4e^{-2t}+\dfrac{9}{2}e^{-3t} 


   3 \alpha-3 \beta+\gamma    & \dfrac{5}{2} \alpha-4 \beta+\dfrac{3}{2} \gamma & \dfrac{1}{2} \alpha-\beta+\dfrac{1}{2} \gamma \\[1em]
  -3 \alpha+6 \beta-3 \gamma  & \dfrac{-5}{2} \alpha+8 \beta-\dfrac{9}{2} \gamma & \dfrac{-1}{2} \alpha+2 \beta-\dfrac{3}{2} \gamma \\[1em]
   3 \alpha-12 \beta+9 \gamma & \dfrac{5}{2} \alpha-16 \beta+\dfrac{27}{2} \gamma & \dfrac{1}{2} \alpha-4 \beta+\dfrac{9}{2} \gamma
where $\alpha=e^{-t}$, $\beta=e^{-2t}$, and $\gamma=e^{-3t}$.


You can always substitute the large symbols or partial expressions with shorter ones. In this case, assuming \alpha, \beta, and \gamma are not used before, you can replace e^{-t}, e^{-2t}, and e^{-3t} by the shorter \alpha, \beta, and \gamma, respectively.

The second approach is to place the equation at the top of the next page followed by an \hrulefill. In this case, the equation will span the two columns. Also, you should introduce the equation some how earlier to let it appear in the right place. The equation is wrapped into a figure* environment with choosing the [!t] specifier for LaTeX to try very hard to place the equation at the top of the next page.

See Page 8 of How to Use the IEEEtran LATEX Class for more details.

| improve this answer | |

I suggest you group the terms e{-t}, e^{-2t}, and e^{-3t} into a column vector named, say, \tilde{e}. This will let you write the matrix as follows (the horizontal lines in the screenshot are there just to illustrated with width of the columns):

enter image description here

Of course, if you can assume that your readers are familiar with matrix algebra notation, this new expression may be written still more compactly...

\newcommand\evec{\tilde{e}}  % choose suitable vector notation
\hrule % just to demonstrate column width

Set $\evec=\begin{bmatrix}e^{-t} & e^{-2t} & e^{-3t} \end{bmatrix}'$. Then $e^{At}$ is given by
\begin{bmatrix} 3 & -3 & 1 \end{bmatrix}\evec & 
\begin{bmatrix} \frac{5}{2} & -4 & \frac{3}{2} \end{bmatrix}\evec &
\begin{bmatrix} \frac{1}{2} & -1 & \frac{1}{2}\end{bmatrix}\evec \, \\[2ex]
\begin{bmatrix} -3 & 6 & -3 \end{bmatrix}\evec & 
\begin{bmatrix} \frac{-5}{2} & 8 & \frac{-9}{2} \end{bmatrix}\evec &
\begin{bmatrix} \frac{-1}{2} & 2 & \frac{-3}{2} \end{bmatrix}\evec \, \\[2ex]
\begin{bmatrix} 3 & -12 & 9 \end{bmatrix}\evec &
\begin{bmatrix} \frac{5}{2} & -16& \frac{27}{2}\end{bmatrix}\evec &
\begin{bmatrix} \frac{1}{2}& -4& \frac{9}{2} \end{bmatrix}\evec \,

\hrule % just to demonstrate column width
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