# Unique Solution

• Jan 22nd 2010, 05:46 AM
Makall
Unique Solution
$
2x_{1} - 6ax_{2} = 3
$

$
3ax_{1} - x_{2} = 3/2
$

The book says the answer is $x_{1} = 3/2(3a+1)$ and $x_{2} = -3/2(3a+1)$ but when I zero out $3ax_{1}$, I get $x_{2} = (-9a-3)/2(9a^2-1)$.
• Jan 22nd 2010, 05:55 AM
Prove It
Quote:

Originally Posted by Makall
$
2x_{1} - 6ax_{2} = 3
$

$
3ax_{1} - x_{2} = 3/2
$

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

Write in matrix form...

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

This is a matrix equation of the form

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

Using some matrix algebra...

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

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

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

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

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

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

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