Solve the system 3x1+2x2−x3=4 x1−2x2+2x3=1 11x1+2x2+x3=14

Question

Solve the system  3x1+2x2−x3=4  x1−2x2+2x3=1  11x1+2x2+x3=14

Damian Roberts · Accepted Answer

Step 1  Given:  3x1+2x2−x3=4    x1−2x2+2x3=1    11x1+2x2+x3=14  Step 2  [32−11−221121][x1x2x3]=[4114]  Therefore augmented matrix is   [32−141−221112114]  Step 3  Row transformations  R2→3R2−R1, R3→3R3−11R1  [32−140−87−10−1614−2]  R3→R3−2R2  [32−140−8770000]  Step 4  Thus,  x3 is free variable  x2=−18(−1−7x3)  x1=14(5−x3)  for x3=1  x2=1  x1=1  Therefore, (111) is one of the solution of given system.

enguinhispi · Accepted Answer

Step 1 Consider the provided question, The given system of equation is, x1−2x2+2x3=1 (I) 3x1+2x2−x3=4 (II) 11x1+2x2+x3=14 (III) Operate (II)−((I)×3) and (III)−((I)×11) we obtain, 8x2−7x3=1 (IV) And, 24x2−21x3=3⇒8x2−7x3=1 (V) From (IV) and (V), 0x2−0x3=0 Thus, the system of equation is written as, x1−2x2+2x3=1 8x2−7x3=1 0x2−0x3=0 And, the corresponding coefficient matrix is, [1−1208−7000] Step 2 Since, the corresponding coefficient matrix in row echelon form, And, the required equivalent system is, x1−2x2+2x3=1 8x2−7x3=1 0x2−0x3=0 Augumented matrix: [1−22108−710000] Since Rank=number of non=zero row in row echelon form of matrix So, rank of coefficient matrix is 2 and, that of augumented matrix is also 2. NKS Therefore, the given system of equation is consistent. Since, number of free variable =number of unknown variable−rank of coefficient matrix =3−2 =1 So, the system has one free variable. let it be x3 and x3=a; a∈R Step 3 x1−2x2+2a=1 8x2−7a=1 0x2−0a=0 ⇒x2=1+7a8 Now,

Solve the system 3x_{1}+2x_{2}-x_{3}=4 x_{1}-2x_{2}+2x_{3}=1 11x_{1}+2x_{2}+x_{3}=14

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