Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right)$$. Then A2025 $-$ A2020 is equal to :
Solution
$$A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right] \Rightarrow {A^2} = \left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$<br><br>$${A^3} = \left[ {\matrix{
1 & 0 & 0 \cr
2 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right] \Rightarrow {A^4} = \left[ {\matrix{
1 & 0 & 0 \cr
3 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$<br><br>$${A^n} = \left[ {\matrix{
1 & 0 & 0 \cr
{n - 1} & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$<br><br>$${A^{2025}} - {A^{2020}} = \left[ {\matrix{
0 & 0 & 0 \cr
5 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right]$$<br><br>$${A^6} - A = \left[ {\matrix{
0 & 0 & 0 \cr
5 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right]$$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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