# Question 9.16: Given the following national income model: Y = C + I0 + G0 C......

Given the following national income model:

$Y = C + I_0 + G_0$

$C = C_0 + b(Y − T)$,      and      $T = T_0 + tY$

where Y = income, C = consumption, T = taxation, $C_0$, b, t, $I_0$, $G_0$ and $T_0$ (autonomous taxation) are constants and $C_0$ > 0; 0 < b < 1; 0 < t < 1:

(a) Write this model as three equations in terms of the variables Y, C and T.
(b) Use Cramer’s rule to derive expressions for the equilibrium level of income, consumption
and taxation.

Step-by-Step
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(a) The variable terms containing Y, C and T are arranged on the LHS of each equation, the
constants on the RHS:

$Y − C = I_0 + G_0$
$−bY + C + bT = C_0$
$−tY + T = T_0$

(b) Using Cramer’s rule, solve for Y, C, T:

$Y={\frac{\Delta_{Y}}{\Delta}}={\frac{I_{0}+\mathrm{G}_{0}+\mathrm{C}_{0} – b\,T_{0}}{1 – b + b t}}$ = the equilibrium level of income

$C=\frac{\Delta_{\mathrm{C}}}{\Delta}=\frac{C_{0}+b\,T_{0} + b(1 – t)(I_{0}+G_{0})}{1 – b + b t}$ = equilibrium level of consumption

$T=\frac{\Delta_{T}}{\Delta}=\frac{T_{0}(1 – b) + t(\mathrm{C}_{0}+I_{0}+\mathrm{G}_{0})}{1 – b + b t}$ equilibrium level of taxation

Since:

$Δ = \begin{vmatrix} 1 & -1 & 0 \\ -b & 1 & b \\ -t & 0 & 1 \end{vmatrix} = (1) \begin{vmatrix}\\ 1 & b \\ 0 & 1 \end{vmatrix} − (−1) \begin{vmatrix} -b & b \\ -t & 1 \end{vmatrix} + (0) = 1 + (−b − (−bt))$

= 1 − b + bt

$Δ_Y = \begin{vmatrix} I_0 + G_0 & -1 & 0 \\ C_0 & 1 & b \\ T_0 & 0 & 1 \end{vmatrix} = (I_0 + G_0) \begin{vmatrix} 1 & b \\ 0 & 1 \end{vmatrix} − (−1) \begin{vmatrix} C_0 & b \\ T_0 & 1 \end{vmatrix} + (0)$

= $(I_0 + G_0) + (C_0 − bT_0)$

$Δ_C = \begin{vmatrix} 1 & I_0 + G_0 & 0 \\ -b & C_0 & b \\ -t & T_0 & 1 \end{vmatrix} = (1) \begin{vmatrix} C_0 & b \\ T_0 & 1 \end{vmatrix} − (I_0 + G_0) \begin{vmatrix} -b & b \\ -t & 1 \end{vmatrix} + (0)$

= $(C_0 − bT_0) − (I_0 + G_0)(−b − (−bt))$

= $C_0 − bT_0 + b(I_0 + G_0)(1 − t)$

$Δ_T = \begin{vmatrix} 1 & -1 & I_0 + G_0 \\ -b & 1 & C_0 \\ -t & 0 & T_0 \end{vmatrix} = (1) \begin{vmatrix} 1 & C_0 \\ 0 & T_0 \end{vmatrix} − (−1) \begin{vmatrix} -b & C_0 \\ -t & T_0 \end{vmatrix} + (I_0 + G_0) \begin{vmatrix} -b & 1 \\ -t & 0 \end{vmatrix}$

= $T_0 − bT_0 + tC_0 + (I_0 + G_0)(0 − (−t))$

= $T_0(1 − b) + t(C_0 + I_0 + G_0)$

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