Let [A = ( pmatrix 1 & 0 & 0 0 & 1 & 1 1 & 0 & 0 pmatrix )].

Question

Let $$A = \left( {\begin{pmatrix} 1 & 0 & 0  \
   0 & 1 & 1  \
   1 & 0 & 0  \ \end{pmatrix} } \right)$$. Then A2025 - A2020 is equal to :

Accepted Answer

$$A = \left[ {\begin{pmatrix} 1 & 0 & 0  \
   0 & 1 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right] \Rightarrow {A^2} = \left[ {\begin{pmatrix} 1 & 0 & 0  \
   1 & 1 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right]$$

$${A^3} = \left[ {\begin{pmatrix} 1 & 0 & 0  \
   2 & 1 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right] \Rightarrow {A^4} = \left[ {\begin{pmatrix} 1 & 0 & 0  \
   3 & 1 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right]$$

$${A^n} = \left[ {\begin{pmatrix} 1 & 0 & 0  \
   {n - 1} & 1 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right]$$

$${A^{2025}} - {A^{2020}} = \left[ {\begin{pmatrix} 0 & 0 & 0  \
   5 & 0 & 0  \
   0 & 0 & 0  \

\end{pmatrix} } \right]$$

$${A^6} - A = \left[ {\begin{pmatrix} 0 & 0 & 0  \
   5 & 0 & 0  \
   0 & 0 & 0  \

\end{pmatrix} } \right]$$

Solutions · 1