Question

1 -3 1 4 - 21 2. Let A= 1 1 3 06 0 1 -1 5-5 2 Do the columns of A span Rº? What does this mean for the general matrix equation Ax = b where be R*? 5 3. (a) Determine if A = 7 homogeneous equation. (8pts) -3 -4 5 2 1 has at least one free variable by solving the 0

Answer 1

-2 1 louw - 0 2 -- > R, R2 - -2 Blu 1 4 Given A= 1 3 0 6 5 -5 4 6 5 -5 3 4 13 0 6 6 3 2 ooo low - - 1 - R₂ R₂ +3R, Ruth, 3 - 14 R 1 R2 - 0 - O 6 6 - 2 D 3 olla O slu va - R,R,- R2 RzR₃-6R2 Rj RG-6R2 0 O O O MG 0 lu Yu O 피 O D -- o 59 78 1 R,+R, + 2 R3 R₂ + R₂ - 1 2 3 R - 0 0 - L 39 O 0 0 80 39

O O O Ray 39 0 - The pivot elements occur in all the four columns. . The colums of A are linearly independent, and form a basis for Re. :. The columus of A span R". . The general matrix equation is consistent for all 5ER", and has a unique solution. 3. Given A= S-3 -4 7 s - The homogeneous equation is Au=0 where TE The augmented matrix is (A !o]~/5 -3 2 1 OO -4 7 5 O -1 1 O Reh 5 2 o 7 S R₂² R₂ + SR, Rez+7R, O -23 - ror O -23 0 - O R₂3 - 1 / R2 1 -7/23 0 O 0 -22 Z - R,R,- 4R2 Rg Ry+23R2 -77²3 0 0 o → n = »u, + 5%=0 is a free variable = 1 / 3 The system has one free variable; . }