1. ## Unique Solution

$\displaystyle 2x_{1} - 6ax_{2} = 3$
$\displaystyle 3ax_{1} - x_{2} = 3/2$

The book says the answer is $\displaystyle x_{1} = 3/2(3a+1)$ and $\displaystyle x_{2} = -3/2(3a+1)$ but when I zero out $\displaystyle 3ax_{1}$, I get $\displaystyle x_{2} = (-9a-3)/2(9a^2-1)$.

2. Originally Posted by Makall
$\displaystyle 2x_{1} - 6ax_{2} = 3$
$\displaystyle 3ax_{1} - x_{2} = 3/2$

The book says the answer is $\displaystyle x_{1} = 3/2(3a+1)$ and $\displaystyle x_{2} = -3/2(3a+1)$ but when I zero out $\displaystyle 3ax_{1}$, I get $\displaystyle x_{2} = (-9a-3)/2(9a^2-1)$.
Write in matrix form...

$\displaystyle \left[\begin{matrix}2 & -6a\\ 3a & -1\end{matrix}\right] \left[\begin{matrix}x_1\\ x_2 \end{matrix}\right] = \left[\begin{matrix} 3 \\ \frac{3}{2}\end{matrix}\right]$

This is a matrix equation of the form

$\displaystyle A\mathbf{x} = \mathbf{b}$.

Using some matrix algebra...

$\displaystyle A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}$

$\displaystyle I\mathbf{x} = A^{-1}\mathbf{b}$

$\displaystyle \mathbf{x} = A^{-1}\mathbf{b}$.

$\displaystyle A^{-1} = \frac{1}{2(-1) - (-6a)3a}\left[\begin{matrix} -1 & 6a \\ -3a & 2 \end{matrix}\right]$

$\displaystyle = \frac{1}{18a^2 - 2}\left[\begin{matrix} -1 & 6a \\ -3a & 2 \end{matrix}\right]$

So $\displaystyle \left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right] = \frac{1}{18a^2 - 2}\left[\begin{matrix} -1 & 6a \\ -3a & 2 \end{matrix}\right] \left[\begin{matrix} 3 \\ \frac{3}{2} \end{matrix}\right]$.

Now simplify and solve for $\displaystyle x_1$ and $\displaystyle x_2$.