1
JEE Main 2026 (Online) 6th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{array}\right]$ and $\mathrm{B}=\left[\mathrm{b}_{i j}\right], 1 \leq i, j \leq 3$. If $\mathrm{B}=\mathrm{A}^{99}-\mathrm{I}$, then the value of $\frac{\mathrm{b}_{31}-\mathrm{b}_{21}}{\mathrm{~b}_{32}}$ is :

A

99

B

199

C

149

D

159

2
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $f: \mathbf{N} \rightarrow \mathbf{Z}$ is defined by

$$ f(n)=\left|\begin{array}{ccc} n & -1 & -5 \\ -2 n^2 & 3(2 k+1) & 2 k+1 \\ -3 n^3 & 3 k(2 k+1) & 3 k(k+2)+1 \end{array}\right|, k \in N, $$

and $\sum\limits_{n=1}^k f(n)=98$, then $k$ is equal to :

A

3

B

4

C

5

D

6

3
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let M be a $3 \times 3$ matrix such that $\mathrm{M}\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \mathrm{M}\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 2\end{array}\right)$ and $\mathrm{M}\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 1 \\ 1\end{array}\right)$. If $\mathrm{M}\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}1 \\ 7 \\ 11\end{array}\right)$, then $x+y+z$ equals :

A

4

B

5

C

7

D

11

4
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let A be a $3 \times 3$ matrix such that

$$ \mathrm{A}^{\mathrm{T}}\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 5 \\ 2 \\ 2 \end{array}\right], \mathrm{A}^{\mathrm{T}}\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 1 \\ 1 \end{array}\right], \mathrm{A}\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \\ 4 \end{array}\right] \text { and } \mathrm{A}\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 1 \\ 3 \\ 1 \end{array}\right] . $$

If $\operatorname{det}(A)=1$, then $\operatorname{det}\left(\operatorname{adj}\left(A^2+A\right)\right)$ is equal to:

A

16

B

25

C

49

D

64

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