JEE Main 2026 — Mathematics Algebra

Let $A = I + C$A=I+C, where $I$I is the identity matrix and $C = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$C=​039​003​000​​.

Calculating the powers of $C$C, we get:

$$C^2 = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$$C2=​039​003​000​​​039​003​000​​=​009​000​000​​

$$C^3 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$C3=​009​000​000​​​039​003​000​​=​000​000​000​​

Since $C^3 = 0$C3=0, all higher powers of $C$C are also zero matrices. Using the binomial expansion for $A^{99}$A99:

$$A^{99} = (I + C)^{99} = I + 99C + ^{99}C_{2}C^2$$A99=(I+C)99=I+99C+99C2​C2

Given $B = A^{99} - I$B=A99−I, we have:

$$B = 99C + 4851C^2$$B=99C+4851C2

Substituting the matrices $C$C and $C^2$C2:

$$B = 99 \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + 4851 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$$B=99​039​003​000​​+4851​009​000​000​​

From this, we can extract the required elements of $B$B:

$$b_{31} = 99 \times 9 + 4851 \times 9 = 9(99 + 4851) = 9 \times 4950 = 44550$$b31​=99×9+4851×9=9(99+4851)=9×4950=44550

$$b_{21} = 99 \times 3 = 297$$b21​=99×3=297

$$b_{32} = 99 \times 3 = 297$$b32​=99×3=297

Now, substituting these values into the given expression:

$$\dfrac{b_{31} - b_{21}}{b_{32}} = \dfrac{44550 - 297}{297} = \dfrac{44550}{297} - 1$$b32​b31​−b21​​=29744550−297​=29744550​−1

$$\dfrac{b_{31} - b_{21}}{b_{32}} = 150 - 1 = 149$$b32​b31​−b21​​=150−1=149

Answer: $149$149