Q40 / 75 · JEE Main 2025 (23 Jan Shift 2) · Mathematics

Let $A=\left[a_{i j}\right]$ be $3 \times 3$ matrix such that $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], A\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$ and $A\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$, then $a_{23}$ equals :

  • Correct answer
    A
    -1
  • B
    2
  • C
    1
  • D
    0

Solution

$\begin{aligned} & \text { Let } A=\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \\ & \therefore\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \\ & \therefore \quad b=0, e=0, h=1 \\ & \text { and }\left[\begin{array}{lll} a & 0 & c \\ d & 0 & f \\ g & 1 & i \end{array}\right]\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right] \\ & \left.\therefore \begin{array}{l} 4 a+3 c=0 \\ 4 d+3 f=1 \\ 4 g+1+3 i=0 \end{array}\right\} ....(1) \end{aligned}$ $\begin{aligned} & \text { and }\left[\begin{array}{ccc}a & 0 & c \\ d & 0 & f \\ g & 1 & i\end{array}\right]\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] \\ & \left.\begin{array}{c}2 a+2 c=1 \\ \therefore 2 d+2 f=0 \\ 2 g+1+2 i=0\end{array}\right\}...(2)\end{aligned}$ From equation (1) and (2) we get $\begin{aligned} & d=1, f=-1 \\ & \therefore \quad a_{23}=-1 \end{aligned}$

Hence, the correct answer is option: 'A'.

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