3 and 4 .Determinants and Matrices
hard

અહી $A=\left[a_{i j}\right]$ એ  $3 \times 3$ શ્રેણિક છે કે જેથી $A \left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], A \left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$ અને $A \left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$, તો  $a _{23}$ ની કિમંત મેળવો.

A$-1$
B$0$
C$2$
D$1$
(JEE MAIN-2025)

Solution

$\begin{array}{l}\text { Let } A=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right] \\ A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] \Rightarrow\left[\begin{array}{l}a_{12} \\ a_{22} \\ a_{32}\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right] \Rightarrow \begin{array}{l}a_{22}=0 ; a_{12}=0 \\ a_{32}=1\end{array} \\ A\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right] \Rightarrow \begin{array}{l}4 a_{11}+a_{12}+3 a_{13}=0 \\ 4 a_{21}+a_{22}+3 a_{25}=1 \Rightarrow 4 a_{21}+3 a_{25}=1 \\ 4 a_{31}+a_{32}+3 a_{33}=0\end{array} \\ A\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] \Rightarrow \begin{array}{l}2 a_{11}+a_{12}+2 a_{13}=1 \\ 2 a_{21}+a_{22}+2 a_{23}=0 \Rightarrow a_{21}+a_{23}=0 \\ 2 a_{31}+a_{32}+2 a_{33}=0 \\ -4 a_{23}+3 a_{23}=1 \Rightarrow a_{23}=-1\end{array}\end{array}$
Standard 12
Mathematics

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