for $k\neq 0, -1, 1$, find the inverse of the matrix

Question

for $k\neq 0, -1, 1$, find the inverse of the matrix

$$\begin{pmatrix} k&0&0\\ 1&k&1\\ -1&1&k \end{pmatrix}$$

how am I supposed to solve this? all I can think of is plugging random numbers in for k but then there could be many different inverses. I don't see how any row operations can turn k into 1 either that I can think of at least.

Answer 1

I seem to be getting something definitive with these row operations:

$R_1-(k-1)R_2\to R_1\\ R_1-R_2\to R_2\\ R_1+R_3\to R_3\\ R_3-R_2\to R_3\\ -\dfrac{1}{k}R_2\to R_2\\ \dfrac{1}{k+1}R_3\to R_3\\ R_3-R_2\to R_2\\ (k-1)R_3+R_2\to R_3\\ R_1+R_3\to R_1\\ \dfrac{1}{k-1}R_3\to R_3\\ R_1-kR_2\to R_1$

To be clear, an operation like $R_1+R_2\to R_3$ means you add the first two rows and replace the third with this sum.

Answer 2

I realize that you are expected to do this using row reduction, but here is an alternate method that you can use to check your work:

The cofactor matrix is given by $C=\begin{bmatrix} k^2-1&-k-1&k+1\\0&k^2&-k\\0&-k&k^2\end{bmatrix}$,

so its transpose is given by $C^{T}=\begin{bmatrix}k^2-1&0&0\\-k-1&k^2&-k\\k+1&-k&k^2\end{bmatrix}$ and therefore

$\displaystyle A^{-1}=\frac{1}{\det(A)}C^{T}=\frac{1}{k(k-1)(k+1)}\begin{bmatrix}k^2-1&0&0\\-k-1&k^2&-k\\k+1&-k&k^2\end{bmatrix}$.