Question

(1 pt) Solve for X [-44 ] [ 1 -10 ] = X AT 1-14 8 40 HINT: Matrix algebra may not be sufficient! (Why not?). X =

Answer 1

Soln we have wepa 4 X I -14 -lo 8 HO a b -lola bll 1 cd ) 3 -40 +46 на -46+4d чь | I - 10 a +361 14 81 dc + 3d -49 +40 +1 40 -14 -46+40-10 TT6 a +357 Hb+ told c +30 hane -49 + 40 +=b. -4b +40 -- +3 b 40-14=d. _4b+8= (+3d Ha tb-4c = 1 atb 4d = -10 ha -d = 14 4b-C - 3d = -8 augmented matrix u given by

Tui -4 01 Ti 1 0 -4 1-lo 14 0 0 -1 14 TO 4 -3.1-8 Peyoum sow- operations to find out reduced son echelon form, | 41- 40] T R3 R₃-R, Li 7 0 -4 | To lo - 4 - 13 TO 4 -1 -3 1-8 Ryt -lo Vy - 10% 7 0 4 | 4 - 13 4 -1 -3 1 TO 0 1 -1 o 0 0 Yo 27/4 -1 2 4 74 L kez 82-8 -41/4 13 -8 -4 0-1 -3 .4 - A / I To 4/27 oT 1/4 -16/27 -41/27 Ret - 1.13 -3 1-8. R2 To 4 -1

41/27 L TR17 R, R Ti O To o o Too -28/27 4/27 112/27 -43/27 t R3 + R2 oloo 427 17/27 -16/27 -4/27 | 310/27 | Rytru- -17/27 -5/27 2+ R3 Jio -28/27 o +/27 0 0 0 0 -43/27 4/27 11/27 - 16/27 14824 -43/112 | 155/56 -17/27 1-52/27 12 1/2 Busty +2883 I 0 0 1 0 0 10 0 0 -14 0 -15/28 -27/14 1 -43/12 155/56 0 43812 A 139/36 R2 y 23 Ryt Ry +48 R2 Ryt 112 Ry 1' 0. o i 0 0 139 7/2" ] -15/20 -29/04 -43/112155/56 ó , 200 R3+ 43 Ry 112 To oo o3 IRAR, + + Ry 10 000 -3 R₂1 R2 tis Ry TO 0 0 0 2 0 0 0 0 -2 So, we have a=3, b =- Thus -3 xel 2 -2 | 2 de-2