JEE Main 2026 — Matrices Question with Solution

Let $A=I+C$, where $I$ is the identity matrix and $C=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]$.

Calculating the powers of $C$, we get:

$C^2=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 9& 0& 0\end{array}\right]$

$C^3=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 9& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

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

$A^{99}=(I+C{)}^{99}=I+99C{+}^{99}{C}_2{C}^2$

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

$B=99C+4851C^2$

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

$B=99\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]+4851\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 9& 0& 0\end{array}\right]$

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

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

$b_{21}=99\times 3=297$

$b_{32}=99\times 3=297$

Now, substituting these values into the given expression:

$\frac{b_{31}-b_{21}}{b_{32}}=\frac{44550-297}{297}=\frac{44550}{297}-1$

$\frac{b_{31}-b_{21}}{b_{32}}=150-1=149$

Answer: $149$