To transform to a matrix, the following guideline has to be adhered.
Y = C + I + G → Y=C+I+G\to Y = C + I + G → Y − C = I + G Y-C=I+G Y − C = I + G
C = a + b Y → C=a+bY\to C = a + bY → C − b Y = a C-bY=a C − bY = a
T = t Y → T=tY\to T = t Y → T − t Y T-tY T − t Y
Y = C + I + G → Y = 12 + 0.8 Y + 100 + 0.1 Y + 300 Y=C+I+G\to Y=12+0.8Y+100+0.1Y+300 Y = C + I + G → Y = 12 + 0.8 Y + 100 + 0.1 Y + 300
Y − 0.9 Y = 412 → Y = 4120 ∴ I = 112 + 0.9 ( 4120 ) Y-0.9Y=412\to Y=4120\therefore I=112+0.9(4120) Y − 0.9 Y = 412 → Y = 4120 ∴ I = 112 + 0.9 ( 4120 ) = 512 =512 = 512
We then replace the values from the given equations where relevant, and placing a coefficient of zero to the missing constants in the guide equation.
1 × Y − 1 × C + 0 × T = I + G 1\times Y-1\times C+0\times T=I+G 1 × Y − 1 × C + 0 × T = I + G
− 0.8 × Y + 1 × C + 0 × T = 12 -0.8\times Y+1\times C + 0\times T=12 − 0.8 × Y + 1 × C + 0 × T = 12
− t × Y + 0 × C + 1 × T = 0 -t\times Y + 0\times C + 1\times T=0 − t × Y + 0 × C + 1 × T = 0
The matrix obtained from the variable coefficients become;
A = A= A = [ 1 − 1 0 − 0.8 1 0 − t − 0 1 ] \begin{bmatrix}
1 & -1 & 0 \\
-0.8 & 1 & 0 \\
-t & -0 & 1 \\
\end{bmatrix} ⎣ ⎡ 1 − 0.8 − t − 1 1 − 0 0 0 1 ⎦ ⎤ X = X= X = [ Y C T ] \begin{bmatrix}
Y \\
C \\
T \\
\end{bmatrix} ⎣ ⎡ Y C T ⎦ ⎤ Z = Z= Z = [ I + G 12 0 ] \begin{bmatrix}
I+G \\
12 \\
0 \\
\end{bmatrix} ⎣ ⎡ I + G 12 0 ⎦ ⎤ but I and G values are known.
I + G = 512 + 300 = 812 I+G=512+300=812 I + G = 512 + 300 = 812
so Z = [ 812 12 0 ] Z=\begin{bmatrix}
812 \\
12 \\
0 \\
\end{bmatrix} Z = ⎣ ⎡ 812 12 0 ⎦ ⎤
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