JEE Mathematics 2026 Question with Solution

Given: $A=\begin{pmatrix}3 & -4\\1 & -1\end{pmatrix}$ and $A^{100}-100B+I=0$.

Find: The sum of all elements of $B^{100}$.

From

$$A^{100}-100B+I=0$$

we get

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

and hence

$$B=\frac{1}{100}(A^{100}+I)$$

the solution next uses the characteristic equation of $A$:

$$|A-\lambda I|=0$$

which gives

$$\lambda^2-2\lambda-1=0$$

So the eigenvalues are

$$\lambda=1\pm\sqrt{2}$$

Using diagonalisation, the provided solution states that the dominant terms cancel symmetrically, and concludes that the sum of all the elements of $B^{100}$ is $2$.

Therefore, the required numerical value is $2$.

Given: $A=\begin{pmatrix}3 & -4\\1 & -1\end{pmatrix}$.

Find: Sum of all elements of $B^{100}$.

The extracted solution provides these steps:

1. Express $B$ using the given matrix equation.
2. Find eigenvalues of $A$ from its characteristic equation.
3. Use diagonalisation to study high powers.

The working shown is:

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

so

$$B=\frac{1}{100}(A^{100}+I)$$

Also,

$$|A-\lambda I|=0$$

leads to

$$\lambda^2-2\lambda-1=0$$

with roots

$$\lambda=1\pm\sqrt{2}$$

The final line of the provided solution states that the required sum is

$$\boxed{2}$$

So the answer is $2$.