Algebra – Matrices – JEE Main 28 Jan 2026 Shift 2

Solution:

Express matrix $A$ as the sum of the identity matrix $I$ and another matrix $M$:
$$A = \begin{bmatrix}3&-4\\ 1&-1\end{bmatrix} = \begin{bmatrix}1&0\\ 0&1\end{bmatrix} + \begin{bmatrix}2&-4\\ 1&-2\end{bmatrix}$$
$$A = I + M, \quad \text{where } M = \begin{bmatrix}2&-4\\ 1&-2\end{bmatrix}$$

Calculate $M^2$:
$$M^2 = \begin{bmatrix}2&-4\\ 1&-2\end{bmatrix} \begin{bmatrix}2&-4\\ 1&-2\end{bmatrix}$$
$$M^2 = \begin{bmatrix}(2)(2) + (-4)(1) & (2)(-4) + (-4)(-2)\\ (1)(2) + (-2)(1) & (1)(-4) + (-2)(-2)\end{bmatrix}$$
$$M^2 = \begin{bmatrix}4 – 4 & -8 + 8\\ 2 – 2 & -4 + 4\end{bmatrix} = \begin{bmatrix}0&0\\ 0&0\end{bmatrix}$$
Since $M^2 = 0$, $M$ is a nilpotent matrix of index 2. Therefore, $M^k = 0$ for all $k \ge 2$.

Expand $A^{100}$ using the binomial theorem, since $I$ and $M$ commute ($IM = MI = M$):
$$A^{100} = (I + M)^{100}$$
$$A^{100} = \binom{100}{0}I^{100} + \binom{100}{1}I^{99}M^1 + \binom{100}{2}I^{98}M^2 + \dots$$

Since $M^2 = 0, M^3 = 0, \dots$, all higher-order terms become zero:
$$A^{100} = I + 100M$$

We are given the equation $A^{100} = 100B + I$:
$$I + 100M = 100B + I$$
$$100M = 100B \Rightarrow B = M$$

We need to find $B^{100}$:
$$B^{100} = M^{100}$$

Since $M^2 = 0$, it follows that $M^{100} = 0$:
$$B^{100} = \begin{bmatrix}0&0\\ 0&0\end{bmatrix}$$

The sum of all the elements of the zero matrix is $0 + 0 + 0 + 0 = 0$.

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