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?
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 $$A^{-1}=\frac{-1}{\lambda^3}\pmatrix{-1&-\lambda&-\lambda^2\\ -\lambda&-\lambda^2&0 \\ -\lambda^2&0&0}$$
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, $$ A^{-1} = B^{-1}P = \pmatrix{\frac{1}{\lambda^3}&\frac{1}{\lambda^2}&\frac{1}{\lambda}\\\frac{1}{\lambda^2}&\frac{1}{\lambda}&0\\\frac{1}{\lambda}&0&0} $$
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.
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.
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} $, $v = \begin{pmatrix} 0 \\ \lambda \end{pmatrix}$, $T = \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 0 \\ \lambda \end{pmatrix} \left(\begin{pmatrix} 0 & \lambda \end{pmatrix} \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 0 \\ \lambda \end{pmatrix} \right)^{-1} = \begin{pmatrix} \frac{1}{\lambda^2} \\ \frac{1}{\lambda} \end{pmatrix}$
$$ \begin{bmatrix} -(v^\top A^{-1} v)^{-1} & T^\top \\ T & A^{-1}-T v^\top A^{-1} \end{bmatrix} = \begin{bmatrix} \begin{pmatrix} \frac{1}{\lambda^3} \end{pmatrix} & \begin{pmatrix} \frac{1}{\lambda^2} & \frac{1}{\lambda} \end{pmatrix} \\ \begin{pmatrix} \frac{1}{\lambda^2} \\ \frac{1}{\lambda} \end{pmatrix} & \begin{pmatrix} \frac{1}{\lambda} & 0 \\ 0 & 0 \end{pmatrix} \end{bmatrix} $$
