Give the solution set of each system 3x2yz1 xyz 2 5x5yz 0Sol

Give the solution set of each system. 3x+2y+z=1 x-y+z= 2 5x+5y-z =0

Solution

(a) We are given that

(1) 3x + 2y + z = 1, (2) x -y + z = 2 and (3) 5x+5y + z = 0 From the 3rd equation, we get z = -5 (x + y). On substituting this value of z in the 1st and the 2nd equations, we get 3x + 2y - 5 (x + y) = 1 or, - 2x-3y = 1 or, 2x + 3y = -1... (4) and x - y - 5 (x + y) = 2 or, -4x - 6y = 2 or 2x+ 3y = -1...(5) Since the equations 4 and 5 are identical, the given systemof equations will have infinitely many solutions and not a unique solution.

(b) We are given that:

(1) x + y - 2z = 0, (2) x - y = -3, (3) 3x- y - 2z = -6 and (4) 2y - 2z = 3 From the 1st equation, we get 2z = x + y or, z = 1/2 (x + y)...(5). From the 4th equation, we get 2z = 2y - 3 or, z = y - 3/2...(6) on equating these two values of z from the 5th and the 6th equations, we get 1/2 (x + y) = y -3/2 or, x + y = 2y -3 or, x = y -3...(7). From the 2nd equation also, we get x = y -3. On substituting z = 1/2 (x + y) (5th equation) in the 3rd equation, we get 3x - y -(x + y) = -6 or, 2x - 2y = -6 or, x - y = -3 or x = y - 3 which is also the same as the 7th equation. Thus the given system of equations will have infinitely many solutions and not a unique solution.

(c) We are given that:

(1) 2x - y - z + w = 4 and (2) x + y + z = -1 These are two equations which, on addition, become 2x - y - z + w + x + y + z  = 4 -1 or, 3x + w = 3 which is a single equation in 2 variables. The given system of equations will have infinitely many solutions and not a unique solution.

(d) We are given that:

(1) x + y -2z = 0, (2) x -y = -3, and (3) 3x -y -2z = 0 From the 1st equation, we get z = 1/2 (x + y) ...(4). From the 3rd equation, we get 2z = 3x -y or, z = 1/2 (3x -y)...(5). On equating the values of z from the 4th and the 5th equations, we get 1/2 (x + y) = 1/2 (3x - y) or, x + y = 3x -y or, 2x = 2y or x = y...(6) From the 2nd equation, we have x - y = -3 or x = y -3... (7) The 6th and the 7th equations are inconsistent. Therefore, the given systemof equations is inconsistent and does not have any solution.

NOTE: Where, the system has infinitely many solutions, we can find many solutions by choosing an arbitrary value of one variable. This will give the corresponding values of the other variables.