Unique Solution

**Makall** · January 22nd 2010, 04:46 AM

$ 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 $x_{2} = (-9a-3)/2(9a^2-1)$ .

**Prove It** · January 22nd 2010, 04:55 AM

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 $x_2$ .

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