D A = 20 – 2 P A + 4 P B + P C S A = 4 P A − 5 D B = 10 + 3 P A − 5 P B + 2 P C S B = 3 P B − 7 D C = 70 + 4 P A + 2 P B − 5 P C S C = 5 P C − 16 D_A= 20 – 2P_A + 4P_B + P_C \\
S_A=4P_A-5 \\
D_B=10+3P_A-5P_B+2P_C \\
S_B=3P_B-7 \\
D_C=70+4P_A+2P_B-5P_C \\
S_C=5P_C-16 D A = 20–2 P A + 4 P B + P C S A = 4 P A − 5 D B = 10 + 3 P A − 5 P B + 2 P C S B = 3 P B − 7 D C = 70 + 4 P A + 2 P B − 5 P C S C = 5 P C − 16
At equilibrium,
D A = S A = > 20 − 2 P A + 4 P B + P C = 4 P A − 5 = > 4 P A + 2 P A − 4 P B − P C = 20 + 5 D_A=S_A \\
=>20-2P_A+4P_B+P_C=4P_A-5 \\
=>4P_A+2P_A-4P_B- P_C=20+5 D A = S A => 20 − 2 P A + 4 P B + P C = 4 P A − 5 => 4 P A + 2 P A − 4 P B − P C = 20 + 5
= > 6 P A − 4 P B − P C = 25 =>6P_A-4P_B- P_C=25 => 6 P A − 4 P B − P C = 25 ....(1)
D B = S B = > 10 + 3 P A − 5 P B + 2 P C = 3 P B – 7 = > − 3 P A + 3 P B + 5 P B − 2 P C = 10 + 7 D_B=S_B \\
=> 10+ 3P_A - 5P_B + 2P_C = 3P_B – 7 \\
=>-3P_A+3P_B+5P_B-2P_C=10+7 \\ D B = S B => 10 + 3 P A − 5 P B + 2 P C = 3 P B –7 => − 3 P A + 3 P B + 5 P B − 2 P C = 10 + 7
= > − 3 P A + 8 P B − 2 P C = 17 =>-3P_A+8P_B-2P_C=17 => − 3 P A + 8 P B − 2 P C = 17 ....(2)
D C = S C D_C = S_C D C = S C
= > 70 + 4 P A + 2 P B − 5 P C = 5 P C − 16 = > − 4 P A − 2 P B + 5 P C + 5 P C = 70 + 16 =>70+4P_A+2P_B-5P_C=5P_C-16 \\
=>-4P_A-2P_B+5P_C+5P_C=70+16 => 70 + 4 P A + 2 P B − 5 P C = 5 P C − 16 => − 4 P A − 2 P B + 5 P C + 5 P C = 70 + 16
= > − 4 P A − 2 P B + 10 P C = 86 =>-4P_A-2P_B+10P_C=86 => − 4 P A − 2 P B + 10 P C = 86 ....(3)
From the system of equations, we get the determinant as,
∣ D ∣ = ∣ 6 − 4 − 1 − 3 8 − 2 − 4 − 2 10 ∣ = 6 ( 80 − 4 ) + 4 ( − 30 − 8 ) − 1 ( 6 + 32 ) = 456 − 152 − 38 = 266 |D|=\begin{vmatrix} 6 &-4 &-1 \\ -3& 8 &-2 \\ -4 &-2 & 10 \end{vmatrix}=6(80-4)+4(-30-8)-1(6+32)=456-152-38=266 ∣ D ∣ = ∣ ∣ 6 − 3 − 4 − 4 8 − 2 − 1 − 2 10 ∣ ∣ = 6 ( 80 − 4 ) + 4 ( − 30 − 8 ) − 1 ( 6 + 32 ) = 456 − 152 − 38 = 266
By Cramer's rule,
P A ∗ = ∣ 25 − 4 − 1 17 8 − 2 86 − 2 10 ∣ ∣ D ∣ = 25 ( 80 − 4 ) + 4 ( 170 + 172 ) − 1 ( − 34 − 688 ) 266 P A ∗ = 25 × 76 + 4 × 342 − 1 ( − 722 ) 266 = 1900 + 1368 + 722 266 = 3990 266 P A ∗ = 15 Q A ∗ = 4 × 15 − 5 = 60 − 5 = 55 P B ∗ = ∣ 6 25 − 1 − 3 17 − 2 − 4 86 10 ∣ ∣ D ∣ = 6 ( 170 + 172 ) − 25 ( − 30 − 8 ) − 1 ( − 258 + 68 ) 266 P B ∗ = 6 × 342 + 25 × 38 − 1 ( − 190 ) 266 = 2052 + 950 + 190 266 = 3192 266 P B ∗ = 12 Q B ∗ = 3 × 12 − 7 = 36 − 7 = 29 P C ∗ = ∣ 6 − 4 25 − 3 8 17 − 4 − 2 86 ∣ ∣ D ∣ = 6 ( 688 + 34 ) + 4 ( − 258 + 68 ) + 25 ( 6 + 32 ) 266 P C ∗ = 6 × 722 − 4 × 190 + 25 × 38 266 = 4332 − 760 + 950 266 = 4522 266 P C ∗ = 17 Q C ∗ = 5 × 17 − 16 = 85 − 16 = 69 D B = 10 + 3 P A − 5 P B + 2 P C d D B d P A = 3 d D B d P B = − 5 d D B d P C = 2 P_A^*=\frac{\begin{vmatrix} 25 &-4 &-1 \\ 17& 8 &-2 \\ 86 &-2 & 10 \end{vmatrix}}{|D|}=\frac{25(80-4)+4(170+172)-1(-34-688)}{266} \\
P_A^*=\frac{25 \times 76+4 \times 342-1(-722)}{266}=\frac{1900+1368+722}{266}=\frac{3990}{266} \\
P_A^*=15 \\
Q_A^*=4 \times 15-5=60-5=55 \\
P_B^*=\frac{\begin{vmatrix} 6 &25 &-1 \\ -3& 17 &-2 \\ -4 &86 & 10 \end{vmatrix}}{|D|}=\frac{6(170+172)-25(-30-8)-1(-258+68)}{266} \\
P_B^*=\frac{6 \times 342+25 \times 38-1(-190)}{266}=\frac{2052+950+190}{266}=\frac{3192}{266} \\
P_B^*=12 \\
Q_B^*=3 \times 12-7=36-7=29 \\
P_C^*=\frac{\begin{vmatrix} 6 &-4 &25 \\ -3& 8 &17 \\ -4 &-2 & 86 \end{vmatrix}}{|D|}=\frac{6(688+34)+4(-258+68)+25(6+32)}{266} \\
P_C^*=\frac{6 \times 722-4 \times 190+25 \times 38}{266}=\frac{4332-760+950}{266}=\frac{4522}{266} \\
P_C^*=17 \\
Q_C^*=5 \times 17-16=85-16=69 \\
D_B=10+3P_A-5P_B+2P_C \\
\frac{\mathrm{d} D_B}{\mathrm{d} P_A}=3 \\
\frac{\mathrm{d} D_B}{\mathrm{d} P_B}=-5 \\
\frac{\mathrm{d} D_B}{\mathrm{d} P_C}=2 P A ∗ = ∣ D ∣ ∣ ∣ 25 17 86 − 4 8 − 2 − 1 − 2 10 ∣ ∣ = 266 25 ( 80 − 4 ) + 4 ( 170 + 172 ) − 1 ( − 34 − 688 ) P A ∗ = 266 25 × 76 + 4 × 342 − 1 ( − 722 ) = 266 1900 + 1368 + 722 = 266 3990 P A ∗ = 15 Q A ∗ = 4 × 15 − 5 = 60 − 5 = 55 P B ∗ = ∣ D ∣ ∣ ∣ 6 − 3 − 4 25 17 86 − 1 − 2 10 ∣ ∣ = 266 6 ( 170 + 172 ) − 25 ( − 30 − 8 ) − 1 ( − 258 + 68 ) P B ∗ = 266 6 × 342 + 25 × 38 − 1 ( − 190 ) = 266 2052 + 950 + 190 = 266 3192 P B ∗ = 12 Q B ∗ = 3 × 12 − 7 = 36 − 7 = 29 P C ∗ = ∣ D ∣ ∣ ∣ 6 − 3 − 4 − 4 8 − 2 25 17 86 ∣ ∣ = 266 6 ( 688 + 34 ) + 4 ( − 258 + 68 ) + 25 ( 6 + 32 ) P C ∗ = 266 6 × 722 − 4 × 190 + 25 × 38 = 266 4332 − 760 + 950 = 266 4522 P C ∗ = 17 Q C ∗ = 5 × 17 − 16 = 85 − 16 = 69 D B = 10 + 3 P A − 5 P B + 2 P C d P A d D B = 3 d P B d D B = − 5 d P C d D B = 2
At equilibrium,
D B = 29 , P A ∗ = 15 , P B ∗ = 12 , P C ∗ = 17 D_B=29 , \small P_A^*=15 , \small P_B^*=12 , \small P_C^*=17 D B = 29 , P A ∗ = 15 , P B ∗ = 12 , P C ∗ = 17
Elasticity of demand with respect to PA,
ε A = d D B d P A P A D B = 3 × 15 29 = 1.55 \varepsilon _A=\frac{\mathrm{d} D_B}{\mathrm{d} P_A}\frac{P_A}{D_B}=3 \times\frac{15}{29}=1.55 ε A = d P A d D B D B P A = 3 × 29 15 = 1.55
ε A > 0 \varepsilon _A>0 ε A > 0 implies A and B are substitute goods.
Elasticity of demand with respect to PB,
ε B = d D B d P B P B D B = − 5 × 12 29 = − 2.0 \varepsilon _B=\frac{\mathrm{d} D_B}{\mathrm{d} P_B}\frac{P_B}{D_B}=-5 \times \frac{12}{29}=-2.0 ε B = d P B d D B D B P B = − 5 × 29 12 = − 2.0
∣ ε B ∣ > 1 |\varepsilon _B|>1 ∣ ε B ∣ > 1 implies the demand for good B is elastic.
Elasticity of demand with respect to PC,
ε C = d D B d P C P C D B = 2 × 17 29 = 1.17 \varepsilon _C=\frac{\mathrm{d} D_B}{\mathrm{d} P_C}\frac{P_C}{D_B}=2 \times\frac{17}{29}=1.17 ε C = d P C d D B D B P C = 2 × 29 17 = 1.17
ε C > 0 \varepsilon _C>0 ε C > 0 implies B and C are substitute goods.
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