Answer to Question #117162 in Linear Algebra for PURBAJYOTI DAS
2020-05-19T06:07:45-04:00
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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2020-05-20T19:32:57-0400
C + I − Y = − 1000 C + 0.7 T − 0.7 Y = 10 I − 0.2 Y = 100 T − 0.3 Y = 0 \begin{alignedat}{2}
C+ I-Y=-1000 \\
C+0.7T-0.7Y =10 \\
I-0.2Y =100 \\
T-0.3Y =0
\end{alignedat} C + I − Y = − 1000 C + 0.7 T − 0.7 Y = 10 I − 0.2 Y = 100 T − 0.3 Y = 0
Δ = ∣ 1 1 0 − 1 1 0 0.7 − 0.7 0 1 0 − 0.2 0 0 1 − 0.3 ∣ = \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}= Δ = ∣ ∣ 1 1 0 0 1 0 1 0 0 0.7 0 1 − 1 − 0.7 − 0.2 − 0.3 ∣ ∣ = = ∣ 0 0.7 − 0.7 1 0 − 0.2 0 1 − 0.3 ∣ − ∣ 1 0 − 1 1 0 − 0.2 0 1 − 0.3 ∣ = =\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 1 0 0.7 0 1 − 0.7 − 0.2 − 0.3 ∣ ∣ − ∣ ∣ 1 1 0 0 0 1 − 1 − 0.2 − 0.3 ∣ ∣ = = − ∣ 0.7 − 0.7 1 − 0.3 ∣ − ∣ 0 − 0.2 1 − 0.3 ∣ + ∣ 0 − 1 1 − 0.3 ∣ = =
-\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}= = − ∣ ∣ 0.7 1 − 0.7 − 0.3 ∣ ∣ − ∣ ∣ 0 1 − 0.2 − 0.3 ∣ ∣ + ∣ ∣ 0 1 − 1 − 0.3 ∣ ∣ = = 0.21 − 0.7 − 0.2 + 1 = 0.31 ≠ 0 =0.21-0.7-0.2+1=0.31\not=0 = 0.21 − 0.7 − 0.2 + 1 = 0.31 = 0
Δ C = ∣ − 1000 1 0 − 1 10 0 0.7 − 0.7 100 1 0 − 0.2 0 0 1 − 0.3 ∣ = \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}= Δ C = ∣ ∣ − 1000 10 100 0 1 0 1 0 0 0.7 0 1 − 1 − 0.7 − 0.2 − 0.3 ∣ ∣ = = − ∣ 10 0.7 − 0.7 100 0 − 0.2 0 1 − 0.3 ∣ − ∣ − 1000 0 − 1 10 0.7 − 0.7 0 1 − 0.3 ∣ = =-\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}= = − ∣ ∣ 10 100 0 0.7 0 1 − 0.7 − 0.2 − 0.3 ∣ ∣ − ∣ ∣ − 1000 10 0 0 0.7 1 − 1 − 0.7 − 0.3 ∣ ∣ = = − 10 ∣ 0 − 0.2 1 − 0.3 ∣ + 100 ∣ 0.7 − 0.7 1 − 0.3 ∣ − =
-10\begin{vmatrix}
0 & -0.2 \\
1 & -0.3
\end{vmatrix}+100\begin{vmatrix}
0.7 & -0.7 \\
1 & -0.3
\end{vmatrix}- = − 10 ∣ ∣ 0 1 − 0.2 − 0.3 ∣ ∣ + 100 ∣ ∣ 0.7 1 − 0.7 − 0.3 ∣ ∣ − − 0.7 ∣ − 1000 − 1 0 − 0.3 ∣ + ∣ − 1000 − 1 10 − 0.7 ∣ = -0.7\begin{vmatrix}
-1000 & -1 \\
0 & -0.3
\end{vmatrix}+\begin{vmatrix}
-1000 & -1 \\
10 & -0.7
\end{vmatrix}= − 0.7 ∣ ∣ − 1000 0 − 1 − 0.3 ∣ ∣ + ∣ ∣ − 1000 10 − 1 − 0.7 ∣ ∣ = = − 2 − 21 + 70 − 210 + 700 + 10 = 547 =-2-21+70-210+700+10=547 = − 2 − 21 + 70 − 210 + 700 + 10 = 547
Δ I = ∣ 1 − 1000 0 − 1 1 10 0.7 − 0.7 0 100 0 − 0.2 0 0 1 − 0.3 ∣ = \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}= Δ I = ∣ ∣ 1 1 0 0 − 1000 10 100 0 0 0.7 0 1 − 1 − 0.7 − 0.2 − 0.3 ∣ ∣ = = ∣ 10 0.7 − 0.7 100 0 − 0.2 0 1 − 0.3 ∣ − ∣ − 1000 0 − 1 100 0 − 0.2 0 1 − 0.3 ∣ = =\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}= = ∣ ∣ 10 100 0 0.7 0 1 − 0.7 − 0.2 − 0.3 ∣ ∣ − ∣ ∣ − 1000 100 0 0 0 1 − 1 − 0.2 − 0.3 ∣ ∣ = = 10 ∣ 0 − 0.2 1 − 0.3 ∣ − 100 ∣ 0.7 − 0.7 1 − 0.3 ∣ + =
10\begin{vmatrix}
0 & -0.2 \\
1 & -0.3
\end{vmatrix}-100\begin{vmatrix}
0.7 & -0.7 \\
1 & -0.3
\end{vmatrix}+ = 10 ∣ ∣ 0 1 − 0.2 − 0.3 ∣ ∣ − 100 ∣ ∣ 0.7 1 − 0.7 − 0.3 ∣ ∣ + + 1000 ∣ 0 − 0.2 1 − 0.3 ∣ + 100 ∣ 0 − 1 1 − 0.3 ∣ = +1000\begin{vmatrix}
0 & -0.2 \\
1 & -0.3
\end{vmatrix}+100\begin{vmatrix}
0 & -1 \\
1 & -0.3
\end{vmatrix}= + 1000 ∣ ∣ 0 1 − 0.2 − 0.3 ∣ ∣ + 100 ∣ ∣ 0 1 − 1 − 0.3 ∣ ∣ = = 2 + 21 − 70 + 200 + 100 = 253 =2+21-70+200+100=253 = 2 + 21 − 70 + 200 + 100 = 253
Δ T = ∣ 1 1 − 1000 − 1 1 0 10 − 0.7 0 1 100 − 0.2 0 0 0 − 0.3 ∣ = \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}= Δ T = ∣ ∣ 1 1 0 0 1 0 1 0 − 1000 10 100 0 − 1 − 0.7 − 0.2 − 0.3 ∣ ∣ = = − 0.3 ∣ 1 1 − 1000 1 0 10 0 1 100 ∣ = =-0.3\begin{vmatrix}
1 & 1 & -1000 \\
1 & 0 & 10 \\
0 & 1 & 100
\end{vmatrix}= = − 0.3 ∣ ∣ 1 1 0 1 0 1 − 1000 10 100 ∣ ∣ = = − 0.3 ∣ 0 10 1 100 ∣ + 0.3 ∣ 1 − 1000 1 100 ∣ = =
-0.3\begin{vmatrix}
0 & 10 \\
1 & 100
\end{vmatrix}+0.3\begin{vmatrix}
1 & -1000 \\
1 & 100
\end{vmatrix}= = − 0.3 ∣ ∣ 0 1 10 100 ∣ ∣ + 0.3 ∣ ∣ 1 1 − 1000 100 ∣ ∣ = = 3 + 30 + 300 = 333 =3+30+300=333 = 3 + 30 + 300 = 333
Δ Y = ∣ 1 1 0 − 1000 1 0 0.7 10 0 1 0 100 0 0 1 0 ∣ = \varDelta_Y=\begin{vmatrix}
1 & 1 & 0 & -1000 \\
1 & 0 & 0.7 & 10 \\
0 & 1 & 0 & 100 \\
0 & 0 & 1 & 0
\end{vmatrix}= Δ Y = ∣ ∣ 1 1 0 0 1 0 1 0 0 0.7 0 1 − 1000 10 100 0 ∣ ∣ = = − ∣ 1 1 − 1000 1 0 10 0 1 100 ∣ = =-\begin{vmatrix}
1 & 1 & -1000 \\
1 & 0 & 10 \\
0 & 1 & 100
\end{vmatrix}= = − ∣ ∣ 1 1 0 1 0 1 − 1000 10 100 ∣ ∣ = = − ∣ 0 10 1 100 ∣ + ∣ 1 − 1000 1 100 ∣ = =
-\begin{vmatrix}
0 & 10 \\
1 & 100
\end{vmatrix}+\begin{vmatrix}
1 & -1000 \\
1 & 100
\end{vmatrix}= = − ∣ ∣ 0 1 10 100 ∣ ∣ + ∣ ∣ 1 1 − 1000 100 ∣ ∣ = = 10 + 100 + 1000 = 1110 =10+100+1000=1110 = 10 + 100 + 1000 = 1110
C = Δ C Δ = 547 0.31 = 54700 31 C={\varDelta_C\over \varDelta}={547\over0.31}={54700\over31} C = Δ Δ C = 0.31 547 = 31 54700
I = Δ I Δ = 253 0.31 = 25300 31 I={\varDelta_I\over \varDelta}={253\over0.31}={25300\over31} I = Δ Δ I = 0.31 253 = 31 25300
T = Δ T Δ = 333 0.31 = 33300 31 T={\varDelta_T\over \varDelta}={333\over0.31}={33300\over31} T = Δ Δ T = 0.31 333 = 31 33300
Y = Δ Y Δ = 1110 0.31 = 111000 31 Y={\varDelta_Y\over \varDelta}={1110\over0.31}={111000\over31} Y = Δ Δ Y = 0.31 1110 = 31 111000
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