Let A= [a_i j ] be 3 × 3 matrix such that A [ … ] and A [ … ], then a_23 equals :

$\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}$From equation (1) and (2) we get$Fromequation(1)and(2)weget\begin{aligned}
& d=1, f=-1 \
& \therefore \quad a_{23}=-1
\end{aligned}$