Answer to Question #117162 in Linear Algebra for PURBAJYOTI DAS

Question #117162
Solve the following national income models by Cramer’s rule Y=C+I+1000,C=10+0.7(Y-T),I=100+0.2Y,T=0.3Y Where Y , C , I, M and T are national income, consumption, investment imports and taxes.
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Expert's answer
2020-05-20T19:32:57-0400
C+IY=1000C+0.7T0.7Y=10I0.2Y=100T0.3Y=0\begin{alignedat}{2} C+ I-Y=-1000 \\ C+0.7T-0.7Y =10 \\ I-0.2Y =100 \\ T-0.3Y =0 \end{alignedat}

Δ=1101100.70.70100.20010.3=\varDelta=\begin{vmatrix} 1 & 1 & 0 & -1 \\ 1 & 0 & 0.7 & -0.7 \\ 0 & 1 & 0 & -0.2 \\ 0 & 0 & 1 & -0.3 \end{vmatrix}==00.70.7100.2010.3101100.2010.3==\begin{vmatrix} 0 & 0.7 & -0.7 \\ 1 & 0 & -0.2 \\ 0 & 1 & -0.3 \end{vmatrix}-\begin{vmatrix} 1 & 0 & -1 \\ 1 & 0 & -0.2 \\ 0 & 1 & -0.3 \end{vmatrix}==0.70.710.300.210.3+0110.3== -\begin{vmatrix} 0.7 & -0.7 \\ 1 & -0.3 \end{vmatrix}-\begin{vmatrix} 0 & -0.2 \\ 1 & -0.3 \end{vmatrix}+\begin{vmatrix} 0 & -1 \\ 1 & -0.3 \end{vmatrix}==0.210.70.2+1=0.310=0.21-0.7-0.2+1=0.31\not=0

ΔC=10001011000.70.7100100.20010.3=\varDelta_C=\begin{vmatrix} -1000 & 1 & 0 & -1 \\ 10 & 0 & 0.7 & -0.7 \\ 100 & 1 & 0 & -0.2 \\ 0 & 0 & 1 & -0.3 \end{vmatrix}==100.70.710000.2010.3100001100.70.7010.3==-\begin{vmatrix} 10 & 0.7 & -0.7 \\ 100 & 0 & -0.2 \\ 0 & 1 & -0.3 \end{vmatrix}-\begin{vmatrix} -1000 & 0 & -1 \\ 10 & 0.7 & -0.7 \\ 0 & 1 & -0.3 \end{vmatrix}==1000.210.3+1000.70.710.3= -10\begin{vmatrix} 0 & -0.2 \\ 1 & -0.3 \end{vmatrix}+100\begin{vmatrix} 0.7 & -0.7 \\ 1 & -0.3 \end{vmatrix}-0.71000100.3+10001100.7=-0.7\begin{vmatrix} -1000 & -1 \\ 0 & -0.3 \end{vmatrix}+\begin{vmatrix} -1000 & -1 \\ 10 & -0.7 \end{vmatrix}==221+70210+700+10=547=-2-21+70-210+700+10=547

ΔI=11000011100.70.7010000.20010.3=\varDelta_I=\begin{vmatrix} 1 & -1000 & 0 & -1 \\ 1 & 10 & 0.7 & -0.7 \\ 0 & 100 & 0 & -0.2 \\ 0 & 0 & 1 & -0.3 \end{vmatrix}==100.70.710000.2010.310000110000.2010.3==\begin{vmatrix} 10 & 0.7 & -0.7 \\ 100 & 0 & -0.2 \\ 0 & 1 & -0.3 \end{vmatrix}-\begin{vmatrix} -1000 & 0 & -1 \\ 100 & 0 & -0.2 \\ 0 & 1 & -0.3 \end{vmatrix}==1000.210.31000.70.710.3+= 10\begin{vmatrix} 0 & -0.2 \\ 1 & -0.3 \end{vmatrix}-100\begin{vmatrix} 0.7 & -0.7 \\ 1 & -0.3 \end{vmatrix}++100000.210.3+1000110.3=+1000\begin{vmatrix} 0 & -0.2 \\ 1 & -0.3 \end{vmatrix}+100\begin{vmatrix} 0 & -1 \\ 1 & -0.3 \end{vmatrix}==2+2170+200+100=253=2+21-70+200+100=253

ΔT=111000110100.7011000.20000.3=\varDelta_T=\begin{vmatrix} 1 & 1 & -1000 & -1 \\ 1 & 0 & 10 & -0.7 \\ 0 & 1 & 100 & -0.2 \\ 0 & 0 & 0 & -0.3 \end{vmatrix}==0.3111000101001100==-0.3\begin{vmatrix} 1 & 1 & -1000 \\ 1 & 0 & 10 \\ 0 & 1 & 100 \end{vmatrix}==0.30101100+0.3110001100== -0.3\begin{vmatrix} 0 & 10 \\ 1 & 100 \end{vmatrix}+0.3\begin{vmatrix} 1 & -1000 \\ 1 & 100 \end{vmatrix}==3+30+300=333=3+30+300=333

ΔY=1101000100.7100101000010=\varDelta_Y=\begin{vmatrix} 1 & 1 & 0 & -1000 \\ 1 & 0 & 0.7 & 10 \\ 0 & 1 & 0 & 100 \\ 0 & 0 & 1 & 0 \end{vmatrix}==111000101001100==-\begin{vmatrix} 1 & 1 & -1000 \\ 1 & 0 & 10 \\ 0 & 1 & 100 \end{vmatrix}==0101100+110001100== -\begin{vmatrix} 0 & 10 \\ 1 & 100 \end{vmatrix}+\begin{vmatrix} 1 & -1000 \\ 1 & 100 \end{vmatrix}==10+100+1000=1110=10+100+1000=1110


C=ΔCΔ=5470.31=5470031C={\varDelta_C\over \varDelta}={547\over0.31}={54700\over31}

I=ΔIΔ=2530.31=2530031I={\varDelta_I\over \varDelta}={253\over0.31}={25300\over31}

T=ΔTΔ=3330.31=3330031T={\varDelta_T\over \varDelta}={333\over0.31}={33300\over31}

Y=ΔYΔ=11100.31=11100031Y={\varDelta_Y\over \varDelta}={1110\over0.31}={111000\over31}


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