Question

Solve by Gaussian elimination with back substitution. (1 - i)X1 + X2X3 - 3= 1-2i (53 +21i) (2 -11i) (-2 + 6i) (35-51) 5+101) +5) (X1, X2, X3) 35 - 5i (b) 3x1+ 1X2+ (1-i)X3= 2+i IX1 4iX2 (-12 +53i) 57 + 8i (24 - 16i) (56 +7i) (x1, X2, X3) - 57+Si (57i) (57+Si) 57 + 8i

Answer 1

solve by onau ssian eli mina ion with ba ck a) (1-i ) 봐 + Sol A ugum enkd Mai s 0 l+ l 0 L-2.

From ③ 2132i 2 ' FYom (i 3--1 1x 5]

3x' tìx2 tい.hx3= 2+; 5 Son Gaussianeli mi na Hon with baek Substhhon Augumen hd Mawix 5+115 0 -1 RN 나; L41 -i 3 14+2; 131.2

FTOM χ2 + (14+21) z3 = 131-2 , 7-56; 56) 子-56 84 +13; 구-5ci Fro m ① 子-56 ; 7-56i 구-56;

구-56; 3 ' 3x, 7-6-150; マー56i 구6-1501 子6-150, 84 +131 子-56i