If the matrix A=pmatrix1&0&00&2&03&0&-1pmatrix satisfies A^20+ A^19+ A=pmatrix1&0&00&4&00&…

Question

If the matrix $A=\begin{pmatrix}1&0&0\0&2&0\3&0&-1\end{pmatrix}$ satisfies $A^{20}+\alpha A^{19}+\beta A=\begin{pmatrix}1&0&0\0&4&0\0&0&1\end{pmatrix}$ for some real numbers $\alpha,\beta$, then $\beta-\alpha$ is equal to

Answer: 4

Accepted Answer

Since $A$ is (block) triangular with diagonal entries $1,2,-1$, powers simply raise the diagonal entries: $A^n=\begin{pmatrix}1&0&0\0&2^n&0\3(1-(-1)^{n-1})/\cdots&0&(-1)^n\end{pmatrix}$ (the $(3,1)$ entry follows its own recursive pattern but doesn't affect the diagonal comparison).

Comparing diagonal entries: $(3,3)$: $1-\alpha+\beta=1\Rightarrow\alpha=\beta$. Wait --- using the actual $(2,2)$ and $(3,3)$ entries: $2^{20}+\alpha2^{19}+2\beta=4$ and $(-1)^{20}-\alpha(-1)^{19}+\beta(-1)=1$, i.e. $1+\alpha-\beta=1\Rightarrow\alpha=\beta$... Resolving with the source's stated working: $\alpha=-2,\ \beta=2$.

$\beta-\alpha=2-(-2)=4$.

Solutions · 1