Question

Consider the system of equations

C) Exercise 2: Consider the system of equations z1 +0z2 +5z3-2x4 +4x5 = 20 x1 +0x2 +23-224 +4x5 = 200 Use MATLAB to answer the following questions (a) Does the system have one solution, multiple solutions, or no solutions? Justify your answer (b) If there is at least one solution, what is the general solution? If no solution exists, you don't need to answer this question.

Answer 1

The augmented matrix of the given system of linear equations is A(say) =

1	0	5	-2	4	20
0	1	1	2	20	10
1	0	1	-2	4	200
1	3	1	-1	4	2

To solve the given system of linear equations, we will reduce A to its RREF as under:

Add -1 times the 1st row to the 3rd row

Add -1 times the 1st row to the 4th row

Add -3 times the 2nd row to the 4th row

Multiply the 3rd row by -1/4

Add 7 times the 3rd row to the 4th row

Multiply the 4th row by -1/5

Add -2 times the 4th row to the 2nd row

Add 2 times the 4th row to the 1st row

Add -1 times the 3rd row to the 2nd row

Add -5 times the 3rd row to the 1st row

Then the RREF of A is

1	0	0	0	28	1951/5
0	1	0	0	-4	-451/5
0	0	1	0	0	-45
0	0	0	1	12	363/5

Thus, the given system of linear equations is equivalent to x₁+28x₅ = 1951/5 or, x₁ = 1951/5-28x₅, x₂ -4x₅ = -451/5 or, x₂ = -451/5+4x₅ , x₃ = -45 and x₄+12x₅ = 363/5 or, x₄ = 363/5-12x₅.

(a). Thus, the given system of linear equations has multiple(infinite) solutions.

(b). The general solution is (x₁,x₂,x₃,x₄,x₅) = (1951/5 -28t, -451/5+4t,-45,363/5 -12t,t) = (1951/5, -451/5,-45,363/5,0) +t(-28, 4,0,-12,1), where x₅ = t is an arbitrary real number.