Let A=[array|c|c| 3 & -4 1 & -1 array] and B be two matrices such that A^100=100B+I.

Question

Let $A=[\begin{array}{|c|c|} 3 & -4 \ 1 & -1 \end{array}]$ and $B$ be two matrices such that ${A}^{100}=100B+I$. Then the sum of all the elements of ${B}^{100}$ is \_\_\_\_\_\_\_

Answer: 0

Accepted Answer

Decompose Matrix A We can write matrix $A$ as the sum of the Identity matrix $I$ and a $C:$

$\begin{gathered} \, A=I+C=[\begin{gathered} 1 \quad 0 \ 0 \quad 1]+[2 \quad -4 \ 1 \quad -2],\, C=[2 \quad -4 \ 1 \quad -2]. \ {C}^{2}=[2 \quad -4 \ 1 \quad -2][2 \quad -4 \ 1 \quad -2]=[4-4 \quad -8+8 \ 2-2 \quad -4+4]=[0 \quad 0 \ 0 \quad 0]=0 \end{gathered}$

Since ${C}^{2}=0,C$ is a nilpotent matrix. Consequently, ${C}^{n}=0$ for all $n\ge 2$.

Find ${A}^{100}$ using Binomial Expansion :

${A}^{100}=(I+C{)}^{100}={}^{100}{C}_{0}{I}^{100}+{}^{100}{C}_{1}{I}^{99}C+{}^{100}{C}_{2}{I}^{98}{C}^{2}+\cdots +{}^{100}{C}_{100}{C}^{100}$

Since ${C}^{2}={C}^{3}=\cdots ={C}^{100}=0$, the expansion simplifies to :

$\begin{gathered} {A}^{100}={}^{100}{C}_{0}I+{}^{100}{C}_{1}C \ {A}^{100}=1\cdot I+100\cdot C \ {A}^{100}=I+100C \end{gathered}$

Find Matrix B :

The problem states that ${A}^{100}=100B+I$. Comparing this to our result:

$\begin{gathered} 100B+I=100C+I \ 100B=100C \ B=C \end{gathered}$

Calculate the Sum of Elements of ${B}^{100}$ :

Since $B=C$ and we know ${C}^{2}=0$, then: $\, {B}^{2}=0$

Consequently, ${B}^{100}=0$ (the zero matrix).

${B}^{100}=[ 0 0 \ 0 0 \end{gathered}]$

The sum of all the elements of ${B}^{100}$ is 0 .

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