Solve Matrices HELP PLEASE?

Question

Determine the equilibrium prices of five independent communities that satisfy:

P1+4P2+2P3-3P4+P5 = 3

-P1-3P2+2P3+P4+2P5 = 4

-2P1-8P2-3P3+2P4+P5 = -7

3P1+12P2+6P3-8P4+2P5 = 9

-P1-4P2-2P3+3P4+2P5 = 3

Answer 1

Let the coefficient matrix A=

[1 4 2 -3 1]

[-1 -3 2 1 2]

[-2 -8 -3 2 1]

[3 12 6 -8 2]

[-1 -4 -2 3 2]

The price vector P=

[P1]

[P2]

[P3]

[P4]

[P5]

And the last vector be e=

[3]

[4]

[-7]

[9]

[3]

The you have

AP = e

and

[A | e] =

[1 4 2 -3 1 3]

[-1 -3 2 1 2 4]

[-2 -8 -3 2 1 -7]

[3 12 6 -8 2 9]

[-1 -4 -2 3 2 3]

First perform (R1 + R2) and (R1 + R5) to get

[1 4 2 -3 1 3]

[0 1 4 -2 3 7]

[-2 -8 -3 2 1 -7]

[3 12 6 -8 2 9]

[0 0 0 0 3 6]

Next do (2R1 + R3), (-3R1 + R4) and R5/3 to get

[1 4 2 -3 1 3]

[0 1 4 -2 3 7]

[0 0 1 -4 3 -1]

[0 0 0 1 -1 0]

[0 0 0 0 1 2]

Now perform (R5 + R4)

[1 4 2 -3 1 3]

[0 1 4 -2 3 7]

[0 0 1 -4 3 -1]

[0 0 0 1 0 2]

[0 0 0 0 1 2]

Now (4R4 +R3) and then (-3R5 + R3)

[1 4 2 -3 1 3]

[0 1 4 -2 3 7]

[0 0 1 0 0 1]

[0 0 0 1 0 2]

[0 0 0 0 1 2]

Now (-4R3 + R2), (2R4 + R2), and then (-3R5 + R2)

[1 4 2 -3 1 3]

[0 1 0 0 0 1]

[0 0 1 0 0 1]

[0 0 0 1 0 2]

[0 0 0 0 1 2]

Now (-4R2 + R1), (-2R3 + R1), (3R4 + R1), and then (-R5 + R1)

[1 0 0 0 0 1]

[0 1 0 0 0 1]

[0 0 1 0 0 1]

[0 0 0 1 0 2]

[0 0 0 0 1 2]

Therefore, the vector P is given by

[1]

[1]

[1]

[2]

[2]

Where P1=1, P2 = 1, P3=1, P4=2, and P5=2