# Thread: Matrix system

1. ## Matrix system

Hey all. a 3 sector model problem, lets jus say not may expertise any help would be great!
Thanks and God bless Bob

2. Originally Posted by bobby87
Hey all. a 3 sector model problem, lets jus say not may expertise any help would be great!
Thanks and God bless Bob
You have the equations:
$Y = C + I^* + G^*$
$C = a(Y - T) + b$
$T = tY + T^*$

You want this in the form
$Ax = B$ <-- I'm using a B to distinguish this from b.

So in the equations put all the Y, C, and T terms on the LHS and everything else on the right:
$Y - C = I^* + G^*$
$C - aY + aT = b$
$T - tY = T^*$

Now arrange the variables Y, C, T in that order in the equations. If a term doesn't exist, give it a 0 coefficient:
$1 \cdot Y - 1 \cdot C + 0 \cdot T = I^* + G^*$
$-aY + 1 \cdot C + aT = b$
$-tY + 0 \cdot C + 1 \cdot T = T^*$

Now you should be able to verify that
$\left ( \begin{matrix} 1 & -1 & 0 \\ -a & 1 & a \\ -t & 0 & 1 \end{matrix} \right ) \cdot \left ( \begin{matrix} Y \\ C \\ T \end{matrix} \right ) = \left ( \begin{matrix} I^* + G^* \\ b \\ T^* \end{matrix} \right )$

-Dan

3. You have the matrix equation:
$\left ( \begin{matrix} 1 & -1 & 0 \\ -a & 1 & a \\ -t & 0 & 1 \end{matrix} \right ) \cdot \left ( \begin{matrix} Y \\ C \\ T \end{matrix} \right ) = \left ( \begin{matrix} I^* + G^* \\ b \\ T^* \end{matrix} \right )$

Cramer's rule says that
$Y = \frac{ \left | \begin{matrix} I^* + G^* & -1 & 0 \\ b & 1 & a \\ T^* & 0 & 1 \end{matrix} \right | }{ \left | \begin{matrix} 1 & -1 & 0 \\ -a & 1 & a \\ -t & 0 & 1 \end{matrix} \right | }$

I'll let you work the determinants. For verification I get that:
$Y = \frac{b + I^* + G^* - aT^*}{a(t - 1) + 1}$

-Dan