Determining $x_2$ in the solution of the system

Question

Task: Determine $x_2$ in the solution of the system$$ \begin{bmatrix}4&a&0\\6&b&2\\9&c&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}1\\2\\0\end{bmatrix}$$when$$ \left|\begin{matrix}4&a&0\\6&b&2\\9&c&3\end{matrix}\right|=4$$ using Cramer's rule.

Options to choose from:

1. $-4$
2. $4$
3. $6$
4. $-2$

My answer: $$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}×4=\begin{bmatrix}1\\2\\0\end{bmatrix}$$

I got $x_2=\dfrac24$ which is not an option to choose. How do I do?

Answer 1

$\underline{\text{Cramer's rule:}}$ $x_i=\dfrac{\Delta_i}{\Delta }$ where $\Delta_i$ is the determinant of matrix $A$ with its $i^{th}$ column replaced by vector $b$ and $\Delta$ is the determinant of the matrix $A$.

$$x_2=\dfrac{\left|\begin{matrix}4 &1 &0\\6 &2 &2\\9 &0 &3\end{matrix}\right|}{4}=\dfrac{24}{4}=6$$