Any trick on finding the inverse of this matrix? Supposing I have a matrix, $\pmatrix{0&0&\lambda\\0&\lambda&-1\\ \lambda&-1&0}$.
Without question you can work out the inverse if this matrix. But since it is highly structured, I suppose there should be some quick way to find out the inverse of it?
Can anyone show me a quick way of thinking about this problem?
 A: The determinant of the matrix is easily seen to be $-\lambda^3$, so Cramer's formula will quickly invert it.
If I've got all of the minors correct, it should be
$$A^{-1}=\frac{-1}{\lambda^3}\pmatrix{-1&-\lambda&-\lambda^2\\ -\lambda&-\lambda^2&0 \\ -\lambda^2&0&0}$$
A: I would have done with a detour: Call the original matrix $A$, we can spot with a permutation matrix $P$ that
$$
PA = \pmatrix{&&1\\&1&\\1&&}A =  \pmatrix{\lambda&-1&0\\ 0&\lambda&-1\\ 0&0&\lambda} =: B
$$
Inverse of this Jordan-like upper triangular $B$ is slightly easier but admittedly requires some affinity: (See Jonas' answer for detailed version)
$$
B^{-1} = \left(\begin{array}{ccc} \frac{1}{\mathrm{\lambda}} & \frac{1}{{\mathrm{\lambda}}^2} & \frac{1}{{\mathrm{\lambda}}^3}\\ 0 & \frac{1}{\mathrm{\lambda}} & \frac{1}{{\mathrm{\lambda}}^2}\\ 0 & 0 & \frac{1}{\mathrm{\lambda}} \end{array}\right)
$$
Finally, 
$$
A^{-1} = B^{-1}P = \pmatrix{\frac{1}{\lambda^3}&\frac{1}{\lambda^2}&\frac{1}{\lambda}\\\frac{1}{\lambda^2}&\frac{1}{\lambda}&0\\\frac{1}{\lambda}&0&0}
$$
A: Let $A$ be your matrix.  Then $J=\pmatrix{0&0&-1\\0&-1&0\\ -1&0&0}A$ (switching the first and third rows and switching signs) is the Jordan block with $-\lambda$ on the diagonal.  You can take the inverse of Jordan blocks using the approach shown in this Wikipedia article on matrix functions.  This yields $J^{-1}=\pmatrix{\frac{-1}{\lambda}&\frac{-1}{\lambda^2}&\frac{-1}{\lambda^3}\\0&\frac{-1}{\lambda}&\frac{-1}{\lambda^2}\\ 0&0&\frac{-1}{\lambda}}$, simply by evaluating $f(-\lambda)$, $f'(-\lambda)$, and $f''(-\lambda)$, where $f(x)=\frac{1}{x}$.  Then note that $A^{-1}=J^{-1}\pmatrix{0&0&-1\\0&-1&0\\ -1&0&0}$, so you switch the first and third columns and switch signs again to get the same result as percusse.
A: This matrix is simple enough that we can find its inverse quite easily by simply writing out the product of it with its inverse:
$$
\pmatrix{0&0&\lambda\\0&\lambda&-1\\ \lambda&-1&0}
\pmatrix{a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}} = 
\pmatrix{1&0&0\\0&1&0\\ 0&0&1}
$$
Taking the first row and first column product, $0a_{11}+0a_{21}+\lambda a_{31} = 1$, so $a_{31} = 1/\lambda$. Now the second row and first column gives us $0a_{11}+\lambda a_{21}+-1/\lambda = 0$, so $a_{21} = 1/\lambda^2$ ... and so on for the rest of the matrix.
A: You can partition the matrix into
$$ \begin{bmatrix} 0 & v^\top \\ v & A \end{bmatrix} $$
where $A$ is a symmetric $2\times2$ matrix and $v$ a $2\times1$ vector, yielding the solution of
$$ \begin{bmatrix} 0 & v^\top \\ v & A \end{bmatrix}^{-1} = \begin{bmatrix} -(v^\top A^{-1} v)^{-1} & T^\top \\ T & A^{-1}-T v^\top A^{-1} \end{bmatrix} $$ 
where $T = A^{-1} v (v^\top A^{-1} v)^{-1} $ is the weighted pseudo inverse of $v$, since $T^\top v=1$ and $v^\top T=1 $
Proof:
$$ \begin{bmatrix} -(v^\top A^{-1} v)^{-1} & T^\top \\ T & A^{-1}-T v^\top A^{-1} \end{bmatrix} \begin{bmatrix} 0 & v^\top \\ v & A \end{bmatrix} = \begin{bmatrix} T^\top v &  -(v^\top A^{-1} v)^{-1} v^\top+T^\top A \\ A^{-1}v-T v^\top A^{-1} v  & T v^\top + A^{-1} A - T v^\top  \end{bmatrix} $$
with the simplification $T^\top A = (v^\top A^{-1} v)^{-1} v^\top$ the above becomes
$  \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $
Example:
$A = \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix} $, $v = \begin{pmatrix} 0 \\ \lambda \end{pmatrix}$, $T = \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 0 \\ \lambda \end{pmatrix} \left(\begin{pmatrix} 0 & \lambda \end{pmatrix}  \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 0 \\ \lambda \end{pmatrix} \right)^{-1} = \begin{pmatrix} \frac{1}{\lambda^2} \\ \frac{1}{\lambda} \end{pmatrix}$
$$ \begin{bmatrix} -(v^\top A^{-1} v)^{-1} & T^\top \\ T & A^{-1}-T v^\top A^{-1} \end{bmatrix} = \begin{bmatrix} 
\begin{pmatrix} \frac{1}{\lambda^3} \end{pmatrix} &
\begin{pmatrix} \frac{1}{\lambda^2} & \frac{1}{\lambda} \end{pmatrix} \\
\begin{pmatrix} \frac{1}{\lambda^2} \\ \frac{1}{\lambda} \end{pmatrix} &
\begin{pmatrix} \frac{1}{\lambda} & 0 \\ 0 & 0 \end{pmatrix}
\end{bmatrix}  $$
