Question

Question complecull Slauus. QUESTION 1 Consider the following system of equations: 411 + 3x2 +213 + 14 = 0 8x1 + 722 +603 +504 = 0 1211 + 11x2 + 1013 +994 = 0 () Find the DIMENSION of the solution space, S.

Answer 1

Given the system of equations : 4x, +3x2+2x3 + x 4 = 0 it is of the 8x, +7x₂+6x₂ + 5x zo form A x = 0 , 12 X, +11 X2 +10x3 +944 70 The coefficient matrin , A= 4 7 8 7 6 5 1 12 11 10 9 First we transform into row reduced form to find the rank of A. (432 ) R₂ = 2R, -R, 14 3 2 1 T 8 7 6 5 10 -1 -2 -3 I 12 11 10 9 R₂ = 3R, -R3 10-2 -4 -6 R₂ = 2R₂ -R3 ( ④ 3 2 1 10 -2 -3) loo.oo There are two pinots element in the row reduced form of A. Sog rank of Aza. ů So, dimension of the solution space s is - A-ramk (A). ( As we know , it A is mxn matrix and rank (A) ar, then dimension of solution space is =n-r =4-2 =2. li ) Now For AX=0 haring solution.

/ 4 3 2 1 10 -1 -2 -3 lo o o o Ooo Una =) A 4x, +3712 + 2x2 + x = 0 -M2-213-314 2 0 --- (1) - - -(2) O + (2) gives an + 2N2-2 nazo =) 2x, + 2₂-1=0. i na ²20,+ 02. 3 Then , 20 z= -₂- 324 = -2-3(2x, +22) =) 2nza -60 - 402 =) az= -30,- 202 So, the general solution X is x=(,,W2,-34,- 202, 22, +42 ) (1,0,-3, 2) + 12 (0, 1, 2, 1) {(1,0,-3, 2), (0, 1, 2, 1) is the required basis. . دو