Mathematics · 2025
JEE Main · 23 January 2025, Shift 2 · Q5
Let A =[ a_ij ] be a 3 × 3 matrix such that A [0; 1; 0] = [0; 0; 1], A [4; 1; 3] = [0; 1; 0] and A [2; 1; 2] = [1; 0; 0], then a_23 equals:
Let $\displaystyle \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]$ be a $\displaystyle 3 \times 3$ matrix such that $\displaystyle \mathrm{A}\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right]$, $\displaystyle \mathrm{A}\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$ and $\displaystyle \mathrm{A}\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$, then $\displaystyle \mathrm{a}_{23}$ equals :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle -1$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.