# Showing Left Side to Right Side.

#### Cynderella

##### New Member
Let $$\mathbf x$$ is a $$(p\times 1)$$ vector, $$\mathbf\mu_1$$ is a $$(p\times 1)$$ vector, $$\mathbf\mu_2$$ is a $$(p\times 1)$$ vector, and $$\Sigma$$ is a $$(p\times p)$$ matrix.

Now I have to show

$$-\frac{1}{2}(\mathbf x-\mathbf\mu_1)'\Sigma^{-1}(\mathbf x-\mathbf\mu_1)+\frac{1}{2}(\mathbf x-\mathbf\mu_2)'\Sigma^{-1}(\mathbf x-\mathbf\mu_2)$$
$$= (\mathbf\mu_1-\mathbf\mu_2)'\Sigma^{-1}\mathbf x-\frac{1}{2}(\mathbf\mu_1-\mathbf\mu_2)'\Sigma^{-1}(\mathbf\mu_1+\mathbf\mu_2)$$

After few lines I got
$$-\frac{1}{2}(\mathbf x-\mathbf\mu_1)'\Sigma^{-1}(\mathbf x-\mathbf\mu_1)+\frac{1}{2}(\mathbf x-\mathbf\mu_2)'\Sigma^{-1}(\mathbf x-\mathbf\mu_2)$$

$$=\frac{1}{2}\mathbf x'\Sigma^{-1}\mathbf\mu_1 + \frac{1}{2}\mathbf\mu_1' \Sigma^{-1}\mathbf x-\frac{1}{2}\mathbf x'\Sigma^{-1}\mathbf\mu_2 - \frac{1}{2}\mathbf\mu_2' \Sigma^{-1}\mathbf x-\frac{1}{2}\mathbf\mu_1' \Sigma^{-1} \mathbf\mu_1 + \frac{1}{2}\mathbf\mu_2' \Sigma^{-1} \mathbf\mu_2$$

Then I have stumbled to reach the final line

$$(\mathbf\mu_1-\mathbf\mu_2)'\Sigma^{-1}\mathbf x-\frac{1}{2}(\mathbf\mu_1-\mathbf\mu_2)'\Sigma^{-1}(\mathbf\mu_1+\mathbf\mu_2).$$