Find basis and dimension {xeR^4|xA=0} where A=[-1,1,2,1,1,0,2,3]^T

Matrix transformations
asked 2021-01-23
Find basis and dimension \(\displaystyle{\left\lbrace{x}{e}{R}^{{4}}{\mid}{x}{A}={0}\right\rbrace}\) where \(\displaystyle{A}={\left[-{1},{1},{2},{1},{1},{0},{2},{3}\right]}^{{T}}\)

Answers (1)

Let S = \(\displaystyle{\left\lbrace{x}∈{R}^{{4}}{\mid}{A}{x}={0}\right\rbrace}\) where A = [-1,1,2,1,1,0,2,3]. Now
PSKS={(x1,x2,x3,x4) ∈R^4: [-1,1,2,1,1,0,2,3](x1,x2,x3,x4)^T=0} ={(x1,x2,x3,x4) ∈R^4: -x+2x2+x3+2x4=0,x1+x2+x4=0} ={(x1,x2,x3,x4) ∈R^4: x3=-3x2-5x4} ={(x1,x2,-3x2-5x4,x4):xi ∈R} ={x1(1,0,0,0)+x2(0,1,-3,0)+x4(0,0,-5,1):xi ∈R} =span{(1,0,0,0),(0,1,-3,0),(0,0,-5,1)}ZSK
Again {(1,0,0,0),(0,1,-3,0),(0,0,-5,1)} is a linearly independent set. Hence {(1,0,0,0),(0,1,-3,0),(0,0,-5,1)} is a basis of S.

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