A have a matrix $M=(a, Ab, A^2b, A^3b)$, where $b=(0,0,0,1)^t$, and an r which is the last row in $M^{-1}$. I know that the matrix M is $$\begin{bmatrix} 0&0&0&1\\0&0&1&-2\\0&1&-2&1\\1&-2&1&1\end{bmatrix}$$ and that $$A=\begin{bmatrix} 0&1&0&0\\0&0&1&0\\0&0&0&1\\-4&-3&-3&-2\end{bmatrix}$$
$$A^2=\begin{bmatrix} 0&0&1&0\\0&0&0&1\\-4&-3&-3&-2\\8&2&3&1\end{bmatrix}$$
$$A^3=\begin{bmatrix} 0&0&0&1\\-4&-3&-3&-2\\8&2&3&1\\-4&5&-1&1\end{bmatrix}$$
and I want to show that the matrix $(r, \ rA, \ rA^2, \ rA^3)^t$ is invertible without using the inverse. How can I do this? I know by checking the inverse that is will generate in I but I don't know how to solve this without using the inverse?
I must say its quite a cool puzzle given the structure of $A$. So first thing one might observe is that the last column of each $A^n$ represents a column in $M$. Therefore, we already know a bit about how $[r; rA; rA^2; rA^3]$ looks like: $$r = [r_1, r_2, r_3, r_4]$$ $$rA = [x_1, x_2, x_3, 0]$$ $$rA^2 = [x_3, x_4, x_5, 0]$$ $$rA^3 = [x_6, x_7, x_8, 1]$$ Furthermore, since $r$ represents the last row of $M^{-1}$, I could also state the following: $$M^{-1} = \begin{bmatrix}c \\ d \\ e \\ r\end{bmatrix}, ~~ \text{such that} \begin{bmatrix}c \\ d \\ e \\ r\end{bmatrix} \begin{bmatrix}b & Ab & A^2b & A^3b\end{bmatrix} = I_{4\times4}$$ Which follows that: $$\begin{bmatrix}rb & rAb & rA^2b & rA^3b\end{bmatrix} = \begin{bmatrix}0 & 0 & 0 & 1\end{bmatrix}$$ Looking at $b$ with respect to $r$, I know that the last value of $r$ must be 0. But if the last value is 0, then the first value of $rA$ is also zero. Additionally, as the last row of $rA$ must be 0 too, the second last value of $r$ must be zero as well. Apply the same logic to $rA^2$ and $rA^3$ and you will finally conclude that $r = [1, 0, 0, 0]$ and that $[r; rA; rA^2; rA^3] = I$.