1. ## matrix equation

Colomn vector

2. Originally Posted by Tom123
Colomn vector x is defined as x = (x1 x2 x3 $)^t$

Solve the matrix equation Ax=b for x1 x2 x3 for:

a = 3 and b = (18 16 62 $)^t$

a = 5 and b = (4 6 24 $)^t$
I don't see "a" in the matrix equation, only "A" and these can't be the same since the given a's are scalars, not matrices.

-Dan

3. Oh sorry... A=

2 0 a
2 2 1
8 a+3 4

4. Originally Posted by Tom123
Colomn vector x is defined as x = (x1 x2 x3 $)^t$

Solve the matrix equation Ax=b for x1 x2 x3 for:

a = 3 and b = (18 16 62 $)^t$

a = 5 and b = (4 6 24 $)^t$
Originally Posted by Tom123
Oh sorry... A=

2 0 a
2 2 1
8 a+3 4
$a = 3$

$b = \left ( \begin{matrix} 18 \\ 16 \\ 62 \end{matrix} \right )$

Solve
$\left ( \begin{matrix} 2 & 0 & 3 \\ 2 & 2 & 1 \\ 8 & 6 & 4 \end{matrix} \right ) \cdot \left ( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right ) = \left ( \begin{matrix} 18 \\ 16 \\ 62 \end{matrix} \right )$

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:
$2x_1 + 3x_3 = 18$
$2x_1 + 2x_2 + x_3 = 16$
$8x_1 + 6x_2 + 4x_3 = 62$

So solve the first equation for $x_3$:
$x_3 = 6 - \frac{2}{3}x_1$
and sub that into the remaining two equations:
$2x_1 + 2x_2 + \left ( 6 - \frac{2}{3}x_1 \right ) = 16 \implies \frac{4}{3}x_1 + 2x_2 = 10$
and
$8x_1 + 6x_2 + 4 \left ( 6 - \frac{2}{3}x_1 \right ) = 62 \implies \frac{16}{3}x_1 + 6x_2 = 38$

Again, solve the top equation, this time for $x_2$:
$x_2 = 5 - \frac{2}{3}x_1$
and sub that into the remaining equation:
$\frac{16}{3}x_1 + 6 \left ( 5 - \frac{2}{3}x_1 \right ) = 38 \implies \frac{4}{3}x_1 = 8$

Now solve for $x_1$:
$x_1 = \frac{3}{4} \cdot 8 = 6$

Thus
$x_2 = 5 - \frac{2}{3} \cdot 6 = 1$
and
$x_3 = 6 - \frac{2}{3} \cdot 6 = 2$

So finally
$\left ( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right ) = \left ( \begin{matrix} 6 \\ 1 \\ 2 \end{matrix} \right )$

You do the other one.

-Dan