Mathematics · Matrices

JEE Main 2026 — 28 January, Evening Shift — Question 25

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

Answer: 0

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{A}=\mathrm{I}+\left[\begin{array}{ll}2 & -4 \\1 & -2\end{array}\right]$ ,

let $\mathrm{M}=\left[\begin{array}{ll}2 & -4 \\1 & -2\end{array}\right]$

$M^{2}=\left[\begin{array}{ll}0 & 0 \\0 & 0\end{array}\right]=M^{3}=M^{4}=\ldots=M^{100}$

$A^{100}=(I+M)^{100}=\sum_{\mathrm{r}=0}^{100}\binom{100}{\mathrm{r}} M^{\mathrm{r}} . \mathrm{I}$

$A^{100}=I+100 M=I+100 B$

$\therefore \mathrm{M}=\mathrm{B} \Rightarrow \mathrm{M}^{100}=\mathrm{B}^{100}=\left[\begin{array}{ll}0 & 0 \\0 & 0\end{array}\right]$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Matrices
Topic: Characteristic Equation & roots,application of cayley - hamilton theorem