JEE Main 2025 Jan 23 Shift 2, Mathematics Q16: Algebra Of Matrices

Each matrix-vector product reads off linear combinations of the columns of $A$.

$A\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$ extracts the second column of $A$: $(a_{12},a_{22},a_{32}{)}^T=(0,0,1{)}^T$. So $a_{12}=0,a_{22}=0,a_{32}=1$.

$A\left[\begin{array}{c}4\\ 1\\ 3\end{array}\right]=(0,1,0{)}^T$ gives the row equations:

Row 2: $4a_{21}+a_{22}+3a_{23}=1\Rightarrow 4a_{21}+3a_{23}=1.$

$A\left[\begin{array}{c}2\\ 1\\ 2\end{array}\right]=(1,0,0{)}^T$ gives:

Row 2: $2a_{21}+a_{22}+2a_{23}=0\Rightarrow 2a_{21}+2a_{23}=0\Rightarrow a_{21}=-a_{23}.$

Substitute into the first equation: $4(-a_{23})+3a_{23}=1\Rightarrow -a_{23}=1\Rightarrow a_{23}=-1.$