Colomn vector
$\displaystyle a = 3$
$\displaystyle b = \left ( \begin{matrix} 18 \\ 16 \\ 62 \end{matrix} \right )$
Solve
$\displaystyle \left ( \begin{matrix} 2 & 0 & 3 \\ 2 & 2 & 1 \\ 8 & 6 & 4 \end{matrix} \right ) \cdot \left ( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right ) = \left ( \begin{matrix} 18 \\ 16 \\ 62 \end{matrix} \right )$
To be precise there are exactly a zillion ways to solve this. I'll assume you don't know any of the fancier ones, and do the matrix multiplication. We get three equations:
$\displaystyle 2x_1 + 3x_3 = 18$
$\displaystyle 2x_1 + 2x_2 + x_3 = 16$
$\displaystyle 8x_1 + 6x_2 + 4x_3 = 62$
So solve the first equation for $\displaystyle x_3$:
$\displaystyle x_3 = 6 - \frac{2}{3}x_1$
and sub that into the remaining two equations:
$\displaystyle 2x_1 + 2x_2 + \left ( 6 - \frac{2}{3}x_1 \right ) = 16 \implies \frac{4}{3}x_1 + 2x_2 = 10$
and
$\displaystyle 8x_1 + 6x_2 + 4 \left ( 6 - \frac{2}{3}x_1 \right ) = 62 \implies \frac{16}{3}x_1 + 6x_2 = 38$
Again, solve the top equation, this time for $\displaystyle x_2$:
$\displaystyle x_2 = 5 - \frac{2}{3}x_1$
and sub that into the remaining equation:
$\displaystyle \frac{16}{3}x_1 + 6 \left ( 5 - \frac{2}{3}x_1 \right ) = 38 \implies \frac{4}{3}x_1 = 8$
Now solve for $\displaystyle x_1$:
$\displaystyle x_1 = \frac{3}{4} \cdot 8 = 6$
Thus
$\displaystyle x_2 = 5 - \frac{2}{3} \cdot 6 = 1$
and
$\displaystyle x_3 = 6 - \frac{2}{3} \cdot 6 = 2$
So finally
$\displaystyle \left ( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right ) = \left ( \begin{matrix} 6 \\ 1 \\ 2 \end{matrix} \right ) $
You do the other one.
-Dan