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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Matrices

Let A=[a_ i j ] be 3 × 3 matrix such that A[ array l 0 1 0 array...

Let A=[aij]A=\left[a_{i j}\right] be 3×33 \times 3 matrix such that A\left[010\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[001\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], A\left[413\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[010\begin{array}{l}0 \\ 1 \\ 0\end{array}\right] and A\left[212\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[100\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], then a23a_{23} equals :

Options

  1. A-1
  2. B2
  3. C1
  4. D0

Correct answer

A. -1

Step-by-step solution

 Let A=[abcdefghi]∴[abcdefghi][010]=[001]∴b=0,e=0,h=1 and [a0cd0fg1i][413]=[010]∴4a+3c=04d+3f=14g+1+3i=0}....(1)\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} and [a0cd0fg1i][212]=[100]2a+2c=1∴2d+2f=02g+1+2i=0}...(2)\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 d=1,f=−1∴a23=−1\begin{aligned} & d=1, f=-1 \\ & \therefore \quad a_{23}=-1 \end{aligned}
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