Question

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6.1. Find the echelon form of the given matrix A, and find its null space N(A) ii. A= /2 4 0 2 3 0 3 3 1 5 2 7 9 7 2 10 0 6 5 3

Answer 1

& Given matrix AF 2 4 0 2 3 To 3 3 1 5 Loo 6 5 3 To find :- 1 Echelon form 2) Null space NCA). Echelon form : A=lo 3 3 1 5 R3-R₃-R1 o 3 3 15 RB R3 R2 2 4 0 23 lo 3 3 5 2 Ruka-D 4 .3 olowo - Loo To % -6 6 5 3 J - According to defination of Row echelon Form; & All zero rows (if any) are at bottom 29 The leading coefficent Cpivot) of non-zero row is always Strictly to right of leading coeffient of above row. Thus to satisfy both condition and is in echelon form. Echelon form is : Toyo2 3 7

9 Null space NCAL & The null space of matrix À are those to with AX20,CO is vector) ie it is solution space for Ar=o. ie Arc= 0 O 3 3 оо 5 2 We can write in augmented form 27 91'2 : 0 from 1 we convert into Row-echelon form Goool 1 5 0 smoo moo 0 ie 2 +42 +224 + 36 20 302 & 3 4g + Ny + 5 x = 0 Borg + 494-6290 - Tq +9X5 = 0 - Here there are 4 equation and 5 variable, so let free variable. Let us at from 9 X = -9X5 3.91 [u= -9 t] one

from ③ blz = - 414 + 6X5 6l = -4694) +6t 63 364 + 64 e 6R₂ = 427 from ② 3X2 = -3% - Tu - 5% =-3 (7t) - [agt) - 54 --214+94-5+ B = -994 from 0; 20 I 20 - 305 12%, 5-4 (1974) - 2 (~94) -37 1.2%, 3 +684 + 18+ -34 12x = +687 +15+ [ 2%, - 113+ Thus, solution space is Cox ma x ma xo]" we can gimplify the above vector; 113 -34 42 -54 677 Tus Null space. NCA span [Mo -34 49-546]"}-