Let $$A = [a_{ij}]$$ be $$3\times 3$$ matrix such that $$A\begin{bmatrix}0 \\1\\0 \end{bmatrix} =\begin{bmatrix}0 \\0\\1 \end{bmatrix},A\begin{bmatrix}4 \\1\\3 \end{bmatrix}=\begin{bmatrix}0 \\1\\0 \end{bmatrix}$$ and $$A\begin{bmatrix}2 \\1\\2 \end{bmatrix}=\begin{bmatrix}1 \\0\\0 \end{bmatrix}$$, then $$a_{23}$$ equals :

We need to find $$a_{23}$$ given three conditions on matrix A.

Write the conditions

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 given conditions:

$$A\begin{bmatrix}0\\1\\0\end{bmatrix} = \begin{bmatrix}0\\0\\1\end{bmatrix}$$ gives us the second column of A: $$a_{12} = 0, a_{22} = 0, a_{32} = 1$$

$$A\begin{bmatrix}4\\1\\3\end{bmatrix} = \begin{bmatrix}0\\1\\0\end{bmatrix}$$ gives:

$$4a_{11} + a_{12} + 3a_{13} = 0$$ ... (i)

$$4a_{21} + a_{22} + 3a_{23} = 1$$ ... (ii)

$$4a_{31} + a_{32} + 3a_{33} = 0$$ ... (iii)

$$A\begin{bmatrix}2\\1\\2\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$$ gives:

$$2a_{11} + a_{12} + 2a_{13} = 1$$ ... (iv)

$$2a_{21} + a_{22} + 2a_{23} = 0$$ ... (v)

$$2a_{31} + a_{32} + 2a_{33} = 0$$ ... (vi)

Find $$a_{23}$$

From (ii): $$4a_{21} + 0 + 3a_{23} = 1 \Rightarrow 4a_{21} + 3a_{23} = 1$$

From (v): $$2a_{21} + 0 + 2a_{23} = 0 \Rightarrow a_{21} = -a_{23}$$

Substituting in (ii): $$4(-a_{23}) + 3a_{23} = 1$$

$$-4a_{23} + 3a_{23} = 1$$

$$-a_{23} = 1$$

$$a_{23} = -1$$

The correct answer is Option 1: -1.