Prove equality powers of matrix

Question

Consider the following matrix $$\begin{bmatrix} 1 & A_{12} & A_{13} & A_{14} \\ 0 & 1 & A_{23} & A_{24} \\ 0&0&1& A_{34}\\0&0&0&1\end{bmatrix}$$

Define $B=A^2, C=A^3, D=A^4$. Prove that $$6B_{14}+D_{14}=4A_{14}+4C_{14}$$

My work so far: I have realised that the characteristic polynomial for this matrix is $(1-\lambda)^4$. I have no clue how to move on, thank you for your explanation!

Answer 1

By Hamilton-Cayley, $(I-A)^4=0$ (since you have already found the char polynomial), so $I-4A+6A^2-4A^3+A^4=0$ or $4A+6C=4B+D+I$, now look at the $(1,4)$-entry and note that $I_{14}=0$.

Answer 2

Hint: Consider the $1,4$ entry of the matrix $$ (6B + D) - (4A + 4C) = A^4 - 4A^3 + 6A^2 - 4 A \\ = [A^4 - 4A^3 + 6A^2 - 4 A + I] - I. $$