Can the matrix $A^{-1}$ be written as a linear combination of $A$ (as given below), where $$A=\begin{bmatrix} a& b& c\\ 0 &a& d\\ 0 &0& a \end{bmatrix}$$ such that $a \neq 0$?
So obviously $\det(A)=a^3 \neq 0$ so $A$ is invertible. Now the dimension of the $3 \times 3$ matrix space over $\mathbb{R}$ is $27$. So $\{I, A, A^2, A^3 \ldots A^{27} \}$ is linearly dependent. So there exists scalars $c_i$ not all zero such that $$c_0I+c_1A+c_2A^2 + \ldots +c_{27}A^{27}=0\,.$$ Now if $c_0 \neq 0$ we can write $$A^{-1}=\frac{1}{c_0}(c_1+c_2A + \ldots +c_{27}A^{26}=0)\,.$$ I cant seem to find a way to guarantee that $c_0 \neq 0$. Need help.
Well, $(A-aI)^3=0$, right? Hence, $$A^3-3aA^2+3a^2A-a^3I=0\,,$$ or $$A^{-1}=a^{-3}A^2-3a^{-2}A+3a^{-1}I\,.$$