Q46 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
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 $\mathrm{B}^{100}$ is $\_\_\_\_$
Correct answer: 0
Solution
$A = I + \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$, let $M = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$
$M^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = M^3 = M^4 = \ldots = M^{100}$
$A^{100} = (I + M)^{100} = \sum_{r=0}^{100} \binom{100}{r} M^r \cdot I$
$A^{100} = I + 100M = I + 100B$
$\therefore M = B \Rightarrow M^{100} = B^{100} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
$M^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = M^3 = M^4 = \ldots = M^{100}$
$A^{100} = (I + M)^{100} = \sum_{r=0}^{100} \binom{100}{r} M^r \cdot I$
$A^{100} = I + 100M = I + 100B$
$\therefore M = B \Rightarrow M^{100} = B^{100} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Hence, the correct answer is 0.