Matrices – Linear Equations – JEE Main 2025 Shift 2

Solution:

Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$.

From the first condition:
$$A\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \Rightarrow \begin{bmatrix} a_{12} \\ a_{22} \\ a_{32} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$
So, $a_{12}=0, a_{22}=0, a_{32}=1$.

From the second condition:
$$A\begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$
Comparing the second row element:
$$4a_{21} + a_{22} + 3a_{23} = 1$$
Since $a_{22}=0$, we get:
$$4a_{21} + 3a_{23} = 1 \quad \dots(1)$$

From the third condition:
$$A\begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$
Comparing the second row element:
$$2a_{21} + a_{22} + 2a_{23} = 0$$
Since $a_{22}=0$, we get:
$$2a_{21} + 2a_{23} = 0 \Rightarrow a_{21} + a_{23} = 0 \Rightarrow a_{21} = -a_{23} \quad \dots(2)$$

Substitute (2) in (1):
$$4(-a_{23}) + 3a_{23} = 1$$
$$-4a_{23} + 3a_{23} = 1$$
$$-a_{23} = 1 \Rightarrow a_{23} = -1$$

Ans. (1)