I have a matrix A and a lower triangular matrix L (with 1's along the diagonal) and an upper triangular matrix U. These are constructed such that $A=LU$. I know that $A^{-1} = L^{-1}U^{-1}$ and I know that the inverse of L is simply the non-diagonal entries with their signs flipped.

Question: Is there an easy way to find the inverse of U?

example: $$\begin{bmatrix}8 & 1 &6\\3 & 5 & 7\\4&9&2\end{bmatrix}^{-1} = \begin{bmatrix}1 & 0 &0\\-.5 & 1 & 0\\-.375 & -.544 & 1\end{bmatrix}\begin{bmatrix}8 & 1 &6\\0 & 8.5 & -1\\0&0&5.294\end{bmatrix}^{-1}$$

I need to find an algorithm for computing the inverse of the far right upper triangular matrix.

  • 6
    $\begingroup$ $A^{-1} = U^{-1} L^{-1}$. Inverse of $L$ is not just the non-diagonal entries with the entries flipped. It is more complicated than that. You can try it for any simple lower triangular matrix. $\endgroup$ Nov 7, 2014 at 9:06

2 Answers 2


A better aproach might be the following.

\begin{equation} \tag{1} \label{inverse} A \cdot A^{-1} = I_3 \end{equation}

Consider $A = L \cdot U$ the $LU$ decomposition of $A$. Then, because of \eqref{inverse},

\begin{equation*} L \cdot U \cdot A^{-1} = I \end{equation*}


\begin{equation*} \begin{bmatrix} 1 & 0 & 0 \\ -.5 & 1 & 0 \\ -.375 & -5.44 & 1 \end{bmatrix} \cdot \begin{bmatrix} 8 & 1 & 6 \\ 0 & 8.5 & -1 \\ 0 & 0 & 5.294 \end{bmatrix} \cdot \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} = \begin{bmatrix} e_1 & e_2 & e_3 \end{bmatrix} \end{equation*}


  • $v_i$ are the colum vectors of $A^{-1}$
  • $e_i$ are standard unit vectors so that the right hand side of the equation represents the identity matrix.

Now you know you can easily calculate $v_i$ in the equation $L \cdot U \cdot v_i = e_i$ for every $i$ and you will have calculated $A^{-1}$.

  • $\begingroup$ Im sorry, but I dont see how this is any easier then calculating the inverse of A in the first place. How can I use my knowledge of L and U in order to make this easier to solve? $\endgroup$ Nov 10, 2014 at 6:52
  • 1
    $\begingroup$ I assume you know how to easily solve $L*U*x = b$. Can you see how you can aply the same tactic for $L*U*v_1 = e_1$,$L*U*v_2 = e_2$ and $L*U*v_3 = e_3$? This way you will have calculated all $v_i$s which make up inverse of $A^{-1}$ $\endgroup$
    – Auberon
    Nov 10, 2014 at 8:25

There is no general "easy" way to compute the inverse of a triangular matrix. Here is one way to do it for a lower triangular matrix. For an upper triangular matrix, you can apply this to take the inverse of its (lower triangular) transpose (which can then be transposed again to give the inverse of the original matrix).


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