This problem requires a clear understanding of matrix-vector multiplication and how to extract specific matrix elements from given conditions. We will focus on the rows of the matrix that contribute to the desired element.

Key Concept: Matrix-Vector Multiplication

For a $3 \times 3$ matrix $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$ and a column vector $X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$, the product $AX$ is a column vector $Y = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}$, where each element $y_i$ is the dot product of the $i$-th row of $A$ with the vector $X$. Specifically, for the second row, the second component of the resulting vector $Y$ is given by: $y_2 = a_{21}x_1 + a_{22}x_2 + a_{23}x_3$ Since we need to find $a_{23}$, we will primarily focus on this relationship for the second row of the matrix $A$.

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Step 1: Use the first condition to find $a_{22}$

The first given condition is $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right]$. Let $X_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ and $Y_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. Applying the matrix-vector multiplication rule for the second row: $a_{21}(0) + a_{22}(1) + a_{23}(0) = y_2$ From $Y_1$, we see that $y_2 = 0$. So, we have: $0 + a_{22} + 0 = 0$ This directly gives us the value of $a_{22}$: $\mathbf{a_{22} = 0}$ This is a crucial piece of information that simplifies our subsequent calculations.

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Step 2: Use the second condition to form an equation for $a_{21}$ and $a_{23}$

The second given condition is $A\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$. Let $X_2 = \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}$ and $Y_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$. Applying the matrix-vector multiplication rule for the second row: $a_{21}(4) + a_{22}(1) + a_{23}(3) = y_2$ From $Y_2$, we see that $y_2 = 1$. Now, substitute the value of $a_{22}=0$ that we found in Step 1: $4a_{21} + (0)(1) + 3a_{23} = 1$ This simplifies to our first linear equation involving $a_{21}$ and $a_{23}$: $\mathbf{4a_{21} + 3a_{23} = 1 \quad \text{(Equation 1)}}$

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Step 3: Use the third condition to form another equation for $a_{21}$ and $a_{23}$

The third given condition is $A\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$. Let $X_3 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ and $Y_3 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$. Applying the matrix-vector multiplication rule for the second row: $a_{21}(2) + a_{22}(1) + a_{23}(2) = y_2$ From $Y_3$, we see that $y_2 = 0$. Substitute the value of $a_{22}=0$: $2a_{21} + (0)(1) + 2a_{23} = 0$ This simplifies to our second linear equation involving $a_{21}$ and $a_{23}$: $2a_{21} + 2a_{23} = 0$ Dividing by 2, we get a simpler relationship: $\mathbf{a_{21} + a_{23} = 0 \quad \text{(Equation 2)}}$

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Step 4: Solve the system of equations for $a_{23}$

We now have a system of two linear equations with two variables ($a_{21}$ and $a_{23}$):

1. $4a_{21} + 3a_{23} = 1$
2. $a_{21} + a_{23} = 0$

From Equation 2, we can easily express $a_{21}$ in terms of $a_{23}$: $a_{21} = -a_{23}$ Now, substitute this expression for $a_{21}$ into Equation 1: $4(-a_{23}) + 3a_{23} = 1$ $-4a_{23} + 3a_{23} = 1$ $-a_{23} = 1$ Multiplying both sides by -1, we find the value of $a_{23}$: $\mathbf{a_{23} = -1}$

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Summary and Key Takeaway

By systematically applying the definition of matrix-vector multiplication to the second row of the matrix $A$ for each given condition, we were able to set up a system of linear equations for $a_{21}$ and $a_{23}$. Solving this system yielded the value of $a_{23}$. This approach highlights how specific elements of a matrix can be determined from its action on particular vectors.

Tips and Common Mistakes:

• Focus on the relevant row/column: Since we needed $a_{23}$ (an element in the second row), we only needed to consider the second row of the matrix $A$ and the second component of the resulting vectors. This saves time and avoids unnecessary calculations.
• Systematic Equation Solving: Be careful when setting up and solving the system of linear equations. A small arithmetic error can lead to a wrong answer.
• Understanding $A \mathbf{e_i}$: The first condition $A\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\\1\end{pmatrix}$ is particularly useful. Multiplying a matrix $A$ by the standard basis vector $\mathbf{e_j}$ (a vector with 1 in the $j$-th position and 0 elsewhere) results in the $j$-th column of $A$. So, $A\mathbf{e_2}$ gives the second column of $A$, which is $\begin{pmatrix}a_{12}\\a_{22}\\a_{32}\end{pmatrix} = \begin{pmatrix}0\\0\\1\end{pmatrix}$. This immediately tells us $a_{12}=0$, $a_{22}=0$, and $a_{32}=1$. This is the quickest way to find $a_{22}$.

The final answer is $\boxed{\text{-1}}$.